A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 444.0 gram setting. It is believed that the machine is underfilling the bags. A 40 bag sample had a mean of 443.0 grams. A level of significance of 0.02 will be used. Determine the decision rule. Assume the standard deviation is known to be 23.0.

Answers

Answer 1

Answer:

We conclude that the  bag filling machine works correctly at the 444.0 gram setting.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 444.0 gram

Sample mean, [tex]\bar{x}[/tex] = 443.0 grams

Sample size, n = 40

Alpha, α = 0.02

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 444.0\text{ grams}\\H_A: \mu < 444.0\text{ grams}[/tex]

We use one-tailed(left) z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{443 - 444}{\frac{23}{\sqrt{40}} } =-0.274[/tex]

Now, [tex]z_{critical} \text{ at 0.02 level of significance } = -2.054[/tex]

Since,  

[tex]z_{stat} < z_{critical}[/tex]

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that the  bag filling machine works correctly at the 444.0 gram setting.

Answer 2
Final answer:

The decision rule for the hypothesis test to check machine calibration involves finding the z-value for a sample mean of 443.0 grams and comparing it with the critical z-value corresponding to a 0.02 significance level for a left-tailed test. If the z-value is less than the critical value, the null hypothesis is rejected.

Explanation:

To determine the decision rule for testing whether the bag filling machine is correctly set at 444.0 grams when we suspect underfilling, we need to perform a hypothesis test using the z-test since the standard deviation is known. Given the sample size (n = 40), sample mean (μ = 443.0 grams), population mean (μ0 = 444.0 grams), population standard deviation (σ = 23.0 grams), and a level of significance of 0.02, we can formulate the null hypothesis (H0: μ = μ0) and the alternative hypothesis (H1: μ < μ0).

The decision rule involves comparing the computed z-value to the critical value from the standard normal distribution at the 0.02 level of significance for a left-tailed test. If the computed z-value is less than the critical z-value, we reject the null hypothesis and accept that the machine is underfilling. The critical z-value is found using the z-table, which corresponds to a cumulative probability of 0.02.


Related Questions

A pediatrician has administered a flu vaccine to 1,000 children in a local community. The average number of children who contract the flu after they get vaccinated is 4 cases per 1,000 children. What is the probability that none of the 1,000 children will contract the flu this season? Hint: Use a Poisson distribution with λ = 4. Group of answer choices

Answers

Answer:

1.83%

Step-by-step explanation:

For the expected value of 4 cases per 1000 children. We can use a formula for Poisson distribution to calculate the probability that none of the 1000 students would contract the flu:

[tex]P(X = 0) = \frac{\lambda^ke^{-\lambda}}{k!}[/tex]

[tex]P(X = 0) = \frac{4^0 e^{-4}}{0!}[/tex]

[tex]P(X = 0) = \frac{1}{e^4} = 0.0183 [/tex]

So the probability of this to happens is 1.83%

Final answer:

The probability that none of the 1,000 children will contract the flu after getting vaccinated, given the average number of flu cases after vaccination is approximately 0.0183 or 1.83%, using the Poisson distribution.

Explanation:

In this question, we are required to use the Poisson distribution to find the probability that none of the 1,000 children will contract the flu this season, given that the average number of cases per 1,000 children after vaccination is 4 (λ = 4).

A Poisson probability can be calculated using the formula P(x; λ) = (e^-λ * λ^x) / x!, where x represents the actual number of successes that result from the experiment, e is the base of the natural logarithm approximated to 2.71828, and λ is the mean number of successes that occur in a specified region.

To find the probability that zero children contract the flu, we use x = 0 in the formula. The Poisson probability P(0; 4) = ((e^-4 * 4^0) / 0!) = e^-4 = 0.0183.

So, the probability that none of the 1,000 children will contract the flu after getting vaccinated is approximately 0.0183 or 1.83%.

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Suppose a 95% confidence interval for the average amount of weight loss on a diet program for males is between 13.0 and 18.0 pounds. These results were based on a sample of 42 male participants who were deemed to be overweight at the start of the 4-month study. What is the margin of error for this study?

Answers

Answer:

The margin of error for this study is 2.5 pounds.

Step-by-step explanation:

The margin of error is the subtraction of the mean by the lower end of the confidence interval, and this must be equal to the subtraction of the upper end to the mean.

In this problem, we have that:

M - 13 = 18 - M

2M = 31

M = 15.5

The mean is 15.5 pounds.

So the margin of error for this study is 15.5 - 13 = 18 - 15.5 = 2.5 pounds.

The test statistic of z equals=2.17 is obtained when testing the claim that pnot equals≠0.2170.217. a. Identify the hypothesis test as being​ two-tailed, left-tailed, or​ right-tailed. b. Find the​ P-value. c. Using a significance level of alphaαequals=0.010.01​, should we reject Upper H 0H0 or should we fail to reject Upper H 0H0​?

Answers

Answer:

a) Two tailed test

Null hypothesis:[tex]p=0.217[/tex]  

Alternative hypothesis:[tex]p \neq 0.217[/tex]  

b) [tex]p_v =2*P(Z>2.17)=0.03[/tex]  

c) If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We Fail to reject the null hypothesis H0

Step-by-step explanation:

Data given and notation

n represent the random sample taken

X represent the outcomes desired in the sample

[tex]\hat p[/tex] estimated proportion of interest

[tex]p_o[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is 0.217 or no:  

a. Identify the hypothesis test as being​ two-tailed, left-tailed, or​ right-tailed.

Two tailed test

Null hypothesis:[tex]p=0.217[/tex]  

Alternative hypothesis:[tex]p \neq 0.217[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

For this case the calculated value is given z =2.17  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

b. Find the​ P-value

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(Z>2.17)=0.03[/tex]  

c. Using a significance level of alphaαequals=0.01, should we reject Upper H 0 or should we fail to reject Upper H 0​?

If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We Fail to reject the null hypothesis H0

Researchers published a study in which they considered the incidence among the elderly of various mental health conditions such as dementia, bi-polar disorder, obsessive compulsive disorder, delirium, and Alzheimer's disease. In the U.S., 45% of adults over 65 suffer from one or more of the conditions considered in the study. Calculate the probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration. Give your answer accurate to three decimal places in decimal form. (Example: 0.398)

Probability =

Answers

Answer:

Probability = 0.100.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Eithey they have one or more of these conditions. Or they do not. This means that the binomial probability distribution is used to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 750, p = 0.45[/tex]

So

[tex]E(X) = 750*0.45 = 337.5[/tex]

[tex]\sqrt{Var(X)} = \sqrt{750*0.45*0.55} = 13.62[/tex]

Calculate the probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration.

This is the pvalue of Z when [tex]X = 320[/tex]. So:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{320 - 337.5}{13.62}[/tex]

[tex]Z = -1.28[/tex]

[tex]Z = -1.28[/tex] has a pvalue of 0.100.

So

Probability = 0.100.

According to Inc, 79% of job seekers used social media in their job search in 2018. Many believe this number is inflated by the proportion of 22- to 30-year-old job seekers who use social media in their job search. Suppose a survey of 22- to 30-year-old job seekers showed that 314 of the 370 respondents use social media in their job search. In addition, 281 of the 370 respondents indicated they have electronically submitted a resume to an employer. (a) Conduct a hypothesis test to determine if the results of the survey justify concluding the proportion of 22- to 30-year-old job seekers who use social media in their job search exceeds the proportion of the population that use social media in their job search. Use α = 0.05. State the null and alternative hypothesis. (Enter != for ≠ as needed.)

Answers

Answer:

Null hypothesis:[tex]p\leq 0.79[/tex]  

Alternative hypothesis:[tex]p > 0.79[/tex]  

[tex]z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/tex]  

[tex]p_v =P(z>2.786)=0.00267[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.  

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=314 represent the respondents that use social media in their job search.

[tex]\hat p=\frac{314}{370}=0.849[/tex] estimated proportion of respondents that use social media in their job search

[tex]p_o=0.79[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is hgiher than 0.79.:  

Null hypothesis:[tex]p\leq 0.79[/tex]  

Alternative hypothesis:[tex]p > 0.79[/tex]  

When we conduct a proportion test we need to use the z statisticc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.786)=0.00267[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.  

A restaurant owner is interested in the proportion of his customers who order dessert. He looks at 65 randomly selected receipts. Match the vocabulary word with its corresponding example.
The proportion of the 65 randomly selected customers who ordered dessert.
The list of the 65 Yes or No answers for whether each customer ordered dessert.
The 65 restaurant patrons whose receipts were observed by the owner.
The proportion of all customers who order dessert.
The answer: Yes or No to whether a customer ordered dessert.
All customers who come to the restaurant.
a. Data
b. Sample
c. Population
d. Statistic
e. Parameter
f. Variable

Answers

Final answer:

The context of a restaurant owner examining dessert orders involves several statistical concepts: Data is the list of Yes or No answers, the Sample is the 65 restaurant patrons, Population is all restaurant customers, the Statistic is the proportion of the sample who ordered dessert, the Parameter is the proportion of all customers who order dessert, and the Variable is whether an individual customer ordered dessert.

Explanation:

The question involves the concepts of statistics in Mathematics, particularly focusing on how different terminologies are used. Here is the matching:

a. Data: This is the list of the 65 Yes or No answers for whether each customer ordered dessert.b. Sample: This applies to the 65 restaurant patrons whose receipts were observed by the owner.c. Population: 'All customers who come to the restaurant' is the complete set, or population in statistical terms.d. Statistic: The proportion of the 65 randomly selected customers who ordered dessert is a statistic because it provides an estimate of a particular characteristic, obtained from the sample.e. Parameter: The proportion of all customers who order dessert is the parameter. It's a numerical characteristic of the population.f. Variable: The answer: Yes or No to whether a customer ordered dessert is the variable in this context, because it can change for each customer or change over time.

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The table below gives the number of hours seven randomly selected students spent studying and their corresponding midterm exam grades. Using this data, consider the equation of the regression line, yˆ = b0 + b1x, for predicting the midterm exam grade that a student will earn based on the number of hours spent studying. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Hours Studying Midterm Grades
1.5 67
2.5 69
3 80
3.5 81
4 86
4.5 89
5.5 94

Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.

Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.

Step 3 of 6: Determine the value of the dependent variable yˆ at x = 0.

Step 4 of 6: Determine if the statement "Not all points predicted by the linear model fall on the same line" is true or false.

Step 5 of 6: Find the estimated value of y when x = 4.5. Round your answer to three decimal places.

Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places

Answers

Answer:

1) [tex]m=\frac{77}{10.5}=7.333[/tex]

2) [tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]

3) [tex]\hat y=7.333(0)+55.190=55.190[/tex]

4) False. The values predited will fall on the same line since we are estimating  the values with just one line.

5) [tex]\hat y=7.333(4.5)+55.190=88.190[/tex]

6) [tex]R^2 = (0.971^2) =0.943[/tex]

And that means that the linear model explains 94.29% of the variation.

Step-by-step explanation:

We assume that the data is this one:

x:1.5,2.5, 3, 3.5 , 4, 4.5 ,5.5

y: 67, 69, 80, 81, 86, 89, 94.

Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

So we can find the sums like this:

[tex]\sum_{i=1}^n x_i = 1.5+2.5+3+3.5+4+4.5+5.5=24.5[/tex]

[tex]\sum_{i=1}^n y_i =67+ 69+ 80+ 81+ 86+ 89+ 94=566[/tex]

[tex]\sum_{i=1}^n x^2_i =1.5^2+2.5^2+3^2+3.5^2+4^2+4.5^2+5.5^2=96.25[/tex]

[tex]\sum_{i=1}^n y^2_i =67^2+69^2+80^2+81^2+86^2+89^2+94^2=46364[/tex]

[tex]\sum_{i=1}^n x_i y_i =1.5*67+2.5*69+3*80+3.5*81+4*86+4.5*89+5.5*94=2058[/tex]

With these we can find the sums:

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=96.25-\frac{24.5^2}{7}=10.5[/tex]

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=2058-\frac{24.5*566}{7}=77[/tex]

And the slope would be:

[tex]m=\frac{77}{10.5}=7.333[/tex]

Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.

Now we can find the means for x and y like this:

[tex]\bar x= \frac{\sum x_i}{n}=\frac{24.5}{7}=3.5[/tex]

[tex]\bar y= \frac{\sum y_i}{n}=\frac{566}{7}=80.857[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]

So the line would be given by:

[tex]\hat y=7.333 x +55.190[/tex]

Step 3 of 6: Determine the value of the dependent variable yˆ at x = 0.

[tex]\hat y=7.333(0)+55.190=55.190[/tex]

Step 4 of 6: Determine if the statement "Not all points predicted by the linear model fall on the same line" is true or false.

False. The values predited will fall on the same line since we are estimating  the values with just one line.

Step 5 of 6: Find the estimated value of y when x = 4.5. Round your answer to three decimal places.

[tex]\hat y=7.333(4.5)+55.190=88.190[/tex]

Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places

n=7 [tex] \sum x = 24.5, \sum y = 566, \sum xy =2058, \sum x^2 =96.25, \sum y^2 =46364[/tex]  

And in order to calculate the correlation coefficient we can use this formula:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

[tex]r=\frac{7(2058)-(24.5)(566)}{\sqrt{[7(96.25) -(24.5)^2][7(46364) -(566)^2]}}=0.971[/tex]

And the determination coeffcient is just the square of the correlation coefficient given by:

[tex]R^2 = (0.971^2) =0.943[/tex]

And that means that the linear model explains 94.3% of the variation.

Final answer:

Regression analysis can be done to find the relationship between hours spent studying and midterm exam grades using the regression line equation yˆ = b0 + b1x. But doing this requires statistical computations and ensuring the correlation coefficient is statistically significant. The coefficient of determination, which indicates the proportion of the variance in the dependent variable predictable from the independent variable, is also needed.

Explanation:

The equation of the regression line you're referring to is written as yˆ = b0 + b1x, where yˆ is the predicted value of the dependent variable (in this case, Midterm Grades), b0 is the y-intercept, b1 is the slope or regression coefficient, and x is the value of the independent variable (Hours Studying). We can use the given data to find the estimated slope and y-intercept, value of yˆ at x = 0, test the truthfulness of the statement, and find the estimated value of y when x = 4.5.

However, being able to do this requires the use of statistical software or methods to calculate corresponding values. Also, this also requires us to know the value of the correlation coefficient to determine if it's statistically significant, as it's not wise to make predictions using a regression line with a correlation coefficient that's not statistically significant.

The last piece to find is the value of the coefficient of determination, which shows the proportion of the dependent variable's variance that's predictable from the independent variable(s). This value also needs computation using formulas or statistical software.

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A statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.81 points, and their standard deviation is 1.06 points. The two-tailed students average 7.64 points, and their standard deviation is 1.33 points.

Answers

Answer:

There is no evidence that there is no significant difference between the sample means

Step-by-step explanation:

given that a  statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.81 points, and their standard deviation is 1.06 points

The two-tailed students average 7.64 points, and their standard deviation is 1.33 points.

Group   One tailed X     Two tailed Y  

Mean 7.8100 7.6400

SD 1.0600 1.3300

SEM 0.1185 0.1487

N 80       80  

[tex]H_0:\bar x=\bar y\\H_a: \bar x \neq \bar y[/tex]

(Two tailed test)

The mean of One tailed X minus Two tailed Y equals 0.1700

t = 0.8940

 df = 158

p value =0.3727

 p is greater than alpha 0.05

There is no evidence that there is no significant difference between the sample means

Listed below are annual data for various years. The data are weights​ (metric tons) of imported lemons and car crash fatality rates per​ 100,000 population. Construct a​ scatterplot, find the value of the linear correlation coefficient​ r, and find the​P-value using α=0.05. Is there sufficient evidence to conclude that there is a linear correlation between lemon imports and crash fatality​ rates? Do the results suggest that imported lemons cause car​ fatalities?

Lemon_Imports_(x) Crash_Fatality_Rate_(y)
230 15.8
264 15.6
359 15.5
482 15.3
531 14.9

1. What are the null and alternative​ hypotheses?
2. Construct a scatterplot.
3. The linear correlation coefficient r is
4. The test statistic t is
5. The​ P-value is

Because the​ P-value is ____ than the significance level 0.05​, there ____ sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.

Do the results suggest that imported lemons cause car​fatalities?

A. The results suggest that an increase in imported lemons causes car fatality rates to remain the same.
B. The results do not suggest any​ cause-effect relationship between the two variables.
C. The results suggest that imported lemons cause car fatalities.
D. The results suggest that an increase in imported lemons causes in an increase in car fatality rates.

Answers

Answer:

Because the​ P-value is _(0.02)  less than the significance level 0.05​, there is sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.

C. The results suggest that imported lemons cause car fatalities.

Step-by-step explanation:

Hello!

The study variables are:

X₁: Weight of imported lemons.

X₂: Car crash fatality rate.

The objective is to test if the imported lemons affect the occurrence of car fatalities. To do so you are asked to use a linear correlation test.

I've made a Scatterplot with the given data, it is attached to the answer.

To be able to use the parametric linear correlation you can use the parametric test (Person) or the nonparametric test Spearman. For Person, you need your variables to have a bivariate normal distribution. Since one of the variables is a discrete variable (ratio of car crashes) and the sample is way too small to make an approximation to a normal distribution, the best test to use is Spearman's rank correlation.

This correlation coefficient (rs) takes values from -1 to 1

If rs = -1 this means that there is a negative correlation between the variables

If rs= 1 this means there is a positive correlation between the variables

If rs =0 then there is no correlation between the variables.

The hypothesis is:

H₀: There is no linear association between X₁ and X₂

H₁: There is a linear association between X₁ and X₂

α: 0.05

To calculate the Spearman's correlation coefficient you have to assign ranks to the observed values of each variable, from the smallest to the highest). Then you have to calculate the difference (d)between the ranks and the square of that difference (d²). (see attachment)

The formula for the correlation coefficient is:

[tex]rs= 1 - \frac{6* (sum of d^2)}{(n-1)n(n+1)}[/tex]

[tex]rs= 1 - \frac{6* (40)}{4*5*6}[/tex]

rs= -1

For this value of the correlation coefficient, the p-value is 0.02

Since the p-value (0.02) is less than the significance level (0.05) the decision is to reject the null hypothesis. In other words, there is a linear correlation between the imported lemons and the car crash fatality ration, this means that the modification in the lemon import will affect the car crash fatality ratio.

Note: the correlation coefficient is negative, so you could say that there is a correlation between the variables and this is negative (meaning that when the lemon import increases, the car crash fatality ratio decreases)

I hope it helps!

A researcher is conducting a chi-square test for independence to evaluate the relationship between gender and preference for three different designs for a new automobile. Each individual in a sample of n=30 males and n=30 females selects a favorite design from the three choices. If the researcher obtains a chi-square statistic of X^2 = 4.81, what is the appropriate statistical decision for the test?
O Fail to reject the null hypothesis with either α = 0.05 or α = 0.01
O There is not enough information to determine the appropriate decision.
O Reject the null hypothesis with α = 0.05 but not with α = 0. 01
O Reject the null hypothesis with either α = 0.05 or α = 0.01

Answers

Answer:

Step-by-step explanation:

You are assigned to the jury of a paternity case; determining whether the the child’s guardian father is actually his biological father. After listening to all the witnesses regarding the child’s family, you are 75% convinced that the guardian father is the child’s biological father. Additionally, you have been presented with laboratory blood tests indicating that the child has blood type B. The laboratory further provided population statistics stating that(a) If the guardian father is assumed to be the biological father, the child has 50% chance of having blood type B.(b) If the guardian father is assumed to NOT be the biological father, the child has 0.91% chance to have a blood type OTHER than B.How confident are you(what is the probability) that the guarding father is the child’s biological father

Answers

Answer:

Answer: 0.6022

Consider the following calculation

Step-by-step explanation:

Let F shows the event that guardian father is biological father. So

P(F) = 0.75

By the complement rule,

P(F') = 1 - P(F) =1 - 0.75 = 0.25

Let B shows the event that child has blood type B. So we have

P(B|F) = 0.50, P(B' |F') = 0.0091

By the complement rule we have

P(B|F') = 1 - P(B' |F') = 0.9909

The probability that the guarding father is the child’s biological father given that child have blood type B is

P(BFPF) P(F|B) = PRI P(BF)P(F) + P(BF)P(F) 0.50 -0.75 0.50 -0.75 +0.9909 - 0.25

0.375 0.622725 = 0.6022

Answer: 0.6022

By determining f prime left parenthesis x right parenthesis equals ModifyingBelow lim With h right arrow 0 StartFraction f left parenthesis x plus h right parenthesis minus f left parenthesis x right parenthesis Over h EndFractionf′(x)=limh→0 f(x+h)−f(x) h​, find f prime left parenthesis 7 right parenthesisf′(7) for the given function. f left parenthesis x right parenthesis equals 5 x squaredf(x)=5x2 f prime left parenthesis 7 right parenthesisf′(7)equals=nothing ​(Simplify your​ answer.)

Answers

Answer:

70 is answer

Step-by-step explanation:

Given that a function in x is

[tex]f(x) = 5x^2[/tex]

we have to find f'(7)

we know by derivative rule derivative of a function is

[tex]f'(x) = lim_({h-->0}) \frac{f(x+h)-f(x)}{h}[/tex]

For finding out at 7 we replace x by 7

[tex]f'(7) = lim_({h-->0}) \frac{f(7+h)-f(7)}{h}[/tex]

=[tex]lim\frac{5(7+h)^2-5*7^2}{h} \\= lim \frac{10h*7+h^2}{h} \\= 70+h = 70[/tex]

So f'(7) = 70

answer is 70

Answer:

f'(7)=70

Step-by-step explanation:

We have the definition of the derivative as:

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]

Now we have a function [tex]f(x)=5x^2[/tex] and we want to approximate the first derivative around x=7, that is [tex]f'(7)[/tex].

We can replace this in the first formula as:

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}= \lim_{h \to 0} \dfrac{5(x+h)^2-5x^2}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{5(x^2+2xh+h^2-x^2)}{h}\\\\f'(x)=\lim_{h \to 0}\dfrac{5(2xh+h^2)}{h}\\\\f'(x)=\lim_{h \to 0}5(2x+h)\\\\f'(x)=10x+lim_{h \to 0}h=10x+0=10x[/tex]

Then, the value for f'(7) is:

[tex]f'(7)=10\cdot 7=70[/tex]

Determine whether the random variable is discrete or continuous. In each case, state the possible values of the random variable.

(a)The number of points scored during a basketball game.
(b) The weight of a T-bone steak.

(A) Is the number of points scored during a basketball game discrete or continuous?

a.The random variable is continuous. The possible value are x=0,1,2,
b.The random variable is discrete. The possible values are x ? 0

Answers

Answer:

a) Discrete

b) continuous

Step-by-step explanation:

Given that there are two random variables

1) The number of points scored during a basketball game

2) The weight of a

T bone steak.

The number of points scored during a basketball game

-- Can take values as 0,1,2.....

This can be counted i.e. we can have a one to one correspondence with natural numbers.

So this is a discrete variable

The weight of a T-bone steak

-- This can take any value in decimal or fraction.  This can be between an interval comprising all values over the interval.  Hence we cannot set one to one correspondence with set of natural numbers.

So continuous variable.

Nathan and Carl are running for the mayor of Middletown, in which 60% of the voters favor Nathan and 40% support Carl. A poll is conducted in which 100 residents, selected at random, are asked their preference. What is the likelihood that the poll will show a majority in favor of Carl?

A. 0.3409
B. 0.0068
C. 0.0207
D. 0.1976

Answers

Answer:

C. 0.0207

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they vote for Carl, or they vote for Nathan. So we use the binomial probability distribution.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]\mu = 100*0.4 = 40[/tex]

[tex]\sigma = \sqrt{100*0.4*0.6} = 4.9[/tex]

What is the likelihood that the poll will show a majority in favor of Carl?

This is 1 subtracted by the pvalue of Z when X = 50. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{50 - 40}{4.9}[/tex]

[tex]Z = 2.04[/tex]

[tex]Z = 2.04[/tex] has a pvalue of 0.9793. So the answer is 1-0.9793 = 0.0207.

Which of the following is the upper critical value of z (z*) for an 80% confidence interval?

a. 1.96
b. .84
c. 2.33
d. 1.45
e. 1.28

Answers

Answer:

e. 1.28

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.8}{2} = 0.1[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex]. This is our critical value.

So it is z with a pvalue of [tex]1-0.1 = 0.9[/tex], so [tex]z = 1.28[/tex]

The correct answer is:

e. 1.28

Nike has decided to sell special high quality insoles for its line of basketball tennis shoes. It has fixed costs of $6 million and unit variable costs of $5 per pair. Nike would like to earn a profit of $2 million; how many pairs must they sell at a price of $15? a. 100,000 kits b. 200,000 kits c. 600,000 kits d. 800,000 kits e. 1,400,000 kits

Answers

Answer:

option (d) 800,000

Step-by-step explanation:

Let the number of pairs that must be sold be 'x'

Thus,

Total variable cost for x units = $5x

Therefore,

Total cost = Fixed cost + Total variable cost

= $6 million + $5x

Now,

Profit = Cost - Revenue

Thus,

$2 million = $15x - ( $6 million + $5x )

or

$2 million + $6 million = $10x

or

$8 million = $10x

or

$8,000,000 = $10x

or

x = 800,000

Hence,

option (d) 800,000

The test statistic of z=2.48 is obtained when testing the claim that p > 0.7.

(a) Identify the hypothesis test as being​ two-tailed, left-tailed, or​ right-tailed.
(b) Find the​ P-value.
(c) Using a significance level of alpha=.05​, should we reject H0 or should we fail to reject H0​?

Answers

Answer:

a)  Right-tailed test

b) 0.006569

c) We reject [tex]H_0[/tex].

Step-by-step explanation:

We are given the following information in the question:

[tex]z_{\text{statistic}} = 2.48[/tex]

We are testing the claim  that p > 0.7

Alpha, α = 0.05

a) We have to identify type of test.

Since we test for claim p > 0.7, it is a one-tailed(right) test.

b) The p-value can be calculated with the help of standard normal table.

[tex]P(z>2.48) = 1 - P(z<2.48) = 0.006569[/tex]

C) Since the p-value is less than the significance level, we fail to accept the null hypothesis and reject the null hypothesis.

Thus, we reject [tex]H_0[/tex].

Final answer:

The hypothesis test in question is right-tailed. The p-value can be obtained from a Z-table or statistical software and is probably less than 0.01. Given a significance level of alpha=.05, we should reject H0 if the p-value is less than alpha.

Explanation:

This question involves the concepts of hypothesis testing, p-value, and significance level which are typically covered in a college-level statistics course.

(a) The test is right-tailed because we are testing the claim that p > 0.7. A right-tailed test looks at whether the test statistic is greater than the hypothesized population parameter.

(b) To find the p-value, you would need to look up the value of z=2.48 in a standard normal (Z) table or use statistical software. The p-value represents the probability that the observed difference occurred by chance assuming the null hypothesis is true. The exact value would depend on your table or software, but generally, a z-score of 2.48 would yield a very small p-value, probably less than 0.01.

(c) If the p-value is less than your significance level (alpha=.05), you would reject the null hypothesis (H0). That happens because the evidence suggests that it is highly unlikely (less than a 5% chance) that the observed difference occurred by chance.

Learn more about Hypothesis Testing here:

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Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from renewable sources) and petrodiesel (from fossil fuel). Random blended fuel samples of size 35 are tested in a lab to ascertain the bio/total carbon ratio (X). If the true (i.e. population) mean is 0.948 and the true (i.e. population) standard deviation is 0.006, what is the distribution of Xbar?

Answers

Answer:

[tex]\bar X \sim N(\mu=0.948, \sigma_{\bar X}=\frac{0.006}{\sqrt{35}}=0.00101)[/tex]

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the bio/total carbon ratio. We know from the problem that the distribution for the parameters for the random variable X are:

[tex]\mu = 0.948[/tex]

[tex]\sigma=0.006[/tex]

We select a sample of n=35 nails. That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

So on this case :

[tex]\bar X \sim N(\mu=0.948, \sigma_{\bar X}=\frac{0.006}{\sqrt{35}}=0.00101)[/tex]

A survey found that 73% of adults have a landline at their residence (event A); 83% have a cell phone (event B). It is known that 2% of adults have neither a cell phone nor a landline. 3. What is the probability that an adult selected at random has both a landline and a cell phone? A. 0.58 B. 0.98 C. 0.6059 D. None of these Work: 4. Given an adult has a cell phone, what is the probability he does not have a landline?
A. 0.27
B. 0.25
C. 0.3012
D. None of these

Answers

Answer:

3. What is the probability that an adult selected at random has both a landline and a cell phone?

A. 0.58

4. Given an adult has a cell phone, what is the probability he does not have a landline?

C. 0.3012

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that an adult has a landline at his residence.

B is the probability that an adult has a cell phone.

C is the probability that a mean is neither of those.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that an adult has a landline but not a cell phone and [tex]A \cap B[/tex] is the probability that an adult has both of these things.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

The sum of all the subsets is 1:

[tex]a + b + (A \cap B) + C = 1[/tex]

2% of adults have neither a cell phone nor a landline.

This means that [tex]C = 0.02[/tex].

73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)

So [tex]A = 0.73, B = 0.83[/tex].

What is the probability that an adult selected at random has both a landline and a cell phone?

This is [tex]A \cap B[/tex].

We have that [tex]A = 0.73[/tex]. So

[tex]A = a + (A \cap B)[/tex]

[tex]a = 0.73 - (A \cap B)[/tex]

By the same logic, we have that:

[tex]b = 0.83 - (A \cap B)[/tex].

So

[tex]a + b + (A \cap B) + C = 1[/tex]

[tex]0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1[/tex]

[tex](A \cap B) = 0.75 + 0.83 - 1 = 0.58[/tex]

So the answer for question 3 is A.

4. Given an adult has a cell phone, what is the probability he does not have a landline?

83% of the adults have a cellphone.

We have that

[tex]b = B - (A \cap B) = 0.83 - 0.58 = 0.25[/tex]

25% of those do not have a landline.

So [tex]P = \frac{0.25}{0.83} = 0.3012[/tex]

The answer for question 4 is C.

A journalist reported that the average amount of time that a French person spends eating lunch at a restaurant is 22 minutes. Perform a hypothesis test to determine if a difference exists between the average time an American spends eating lunch when compared to a person from France. The following data represents the​ time, in​ minutes, that random French and American diners spent at lunch. Assume that the population variances are equal. Assume Population 1 is defined as French diners and Population 2 is defined as American diners. What is the test statistic for this hypothesis​ test?

American

21

17

17

20

25

16

20

16

French

24

18

20

28

18

29

17

Answers

Answer:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]

Step-by-step explanation:

Data given

American: 21,17,17,20,25,16,20,16 (Sample 1)

French: 24,18,20,28,18,29,17 (Sample 2)

When we have two independent samples from two normal distributions with equal variances we are assuming that  

[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 = \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]

Our notation on this case :

[tex]n_1 =8[/tex] represent the sample size for group 1

[tex]n_2 =7[/tex] represent the sample size for group 2

[tex]\bar X_1 =19[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =22[/tex] represent the sample mean for the group 2

[tex]s_1=3.117[/tex] represent the sample standard deviation for group 1

[tex]s_2=5.0[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]S^2_p =\frac{(8-1)(3.117)^2 +(7 -1)(5.0)^2}{8 +7 -2}=16.770[/tex]

And the deviation would be just the square root of the variance:

[tex]S_p=4.095[/tex]

And now we can calculate the statistic:

[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=8+7-2=13[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =2*P(t_{13}<-1.416) =0.1803[/tex]

So with the p value obtained and using the significance level assumed [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance we don't have significant differences between the two means.  

Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline that will have the best fuel economy. The four grades are: regular, below regular, premium, and super premium. The test car made three trial runs on the test track using each of the four grades and the miles per gallon recorded. At the 0.05 level, what is the critical value of F used to test the hypothesis that the miles per gallon for each fuel is the same. Show your work.


Miles Per Gallon

Regular: Below Regular Premium Super Premium

$39.31 36.69 38.99 40.04

39.87 40.00 40.02 39.89

39.87 41.01 39.99 39.93

A.) 3.49
B.) 4.07
C.) 2.33
D.) 3.86
E.) 3.26

Answers

Answer:

B) 4.07

Step-by-step explanation:

First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:

Regular    BelowRegular   Premium   SuperPremium

39.31             36.69                38.99             40.04

39.87            40.00                40.02             39.89

39.87            41.01                  39.99             39.93

X1⁻=39.68    X2⁻= 39.23       X3⁻= 39.66    X4⁻=  39.95

Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63

Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):

SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12

SSW =  11.12

Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom

SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974

SSB =  0.80

For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom

Now we can establish the hypothesis for the test:

H0: μ1 = μ2 = μ3 = μ4

The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:

H1: the grade of gasoline does makes a difference

We will use the F statistic to test the hypothesis, which is calculated like follows:

F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19

We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:

Fc = 4.07

Final answer:

The critical value of F used to test the hypothesis that the miles per gallon for each fuel is the same is 3.49.

Explanation:

To test the hypothesis that the miles per gallon for each fuel is the same, we can use an ANOVA test. The critical value of F at the 0.05 level can be found using the F-distribution table or by using statistical software. Since we have three trial runs for each grade of gasoline, we have a total of 12 observations. At a significance level of 0.05 and with 3 degrees of freedom for the numerator and 8 degrees of freedom for the denominator, the critical value of F is 3.49.

A symbol used to name one or more parts of a whole or a set, or a location on a number line is a?​

Answers

Answer:

Fraction

Step-by-step explanation:

Fraction is a symbol that represents a part of a whole. It consists of a numerator and a denominator. The numerator is the number above the fraction bar (also known as "Vinculum), while the denominator is the number below the fraction bar. The denominator is the total number of equal parts in a whole.

Examples of Fraction: [tex]\frac{1}{2}[/tex], [tex]\frac{2}{7}[/tex] and [tex]\frac{5}{8}[/tex].

In the first example, [tex]\frac{1}{2}[/tex], 1 is the numerator, while 2, is the denominator.

Additional Information

When the numerator is smaller than the denominator, the fraction is called "proper fraction." On the contrary, when the numerator is bigger than the denominator, the fraction is called "improper fraction."

Zak is ordering custom T-shirts for his soccer team. Long-sleeved shirts cost $15 each and short-sleeved shirts cost $10 each. Jacob can spend at most $270 and he wants to order at least 20 shirts. Write the constraints using inequalities.

Question 1 options:

15x + 10y < 270


15x + 10y ≤ 270


15x + 10y ≥ 270


x + y ≥ 20


x + y ≤ 20


x > 0


x ≥ 0


y < 0


y ≥ 0

Answers

15x+10y≤270 and x+y≥20 are the constraints for this situation.

Step-by-step explanation:

Given,

Cost of long sleeved shirt = $15

Cost of short sleeved shirt = $10

Amount to spend = $270

Shirts to order = 20

Let,

Long sleeved shirts = x

Short sleeved shirts = y

At most means he can not spend more than $270, therefore,

15x+10y≤270

At least 20 means, he needs minimum 20 or more, therefore,

x+y≥20

15x+10y≤270 and x+y≥20 are the constraints for this situation.

Keywords: inequality, addition

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For the Sequence 2,6,10,14. Write the recursive rule, and a explicit rule

Answers

Answer:

The recursive formula for given series is: [tex]a_{n} = a_{n-1} + 4[/tex]

The explicit formula is given by: [tex]a_{n} = 10 + 5(n-1)[/tex]

Step-by-step explanation:

Determining the Recursive Formula:

As the given sequence is:  2,6,10,14The difference d can be computed by taking the difference between the consecutive terms of the mentioned sequence.

         d = 6 -2 = 10 - 6 = 14 - 10 = 4

It is observed that the difference between the consecutive terms remains constant) with common difference d = 4We know that if the difference between the consecutive terms remains constant), then the series is in arithmetic series. The recursive formula is: [tex]a_{n} = a_{n-1} + d[/tex]So, the recursive formula for given series is: [tex]a_{n} = a_{n-1} + 4[/tex]

Determining the Explicit  Rule or Formula:

An explicit formula defines the nth term of the sequence, where n being the term's location. In other words, a sequence can be defined as a formula in  terms of n. So,

First determine the sequence whether the sequence is in arithmetic. As we know that 2,6,10,14 is in arithmetic. Then find the common difference. Here, d = 6 -2 = 10 - 6 = 14 - 10 = 4Establishing an explicit formula by analyzing the pattern. i.e. adding first term to the product of d (common difference) and one less than the term number

Hence, the explicit formula is given by:

[tex]a_{n} = a_{1} + d (n-1)[/tex]

where,

aₙ is the nth term of the sequence

n is the term number

a₁ is the first term

d is the common difference

As the given sequence is:  2,6,10,14

a₁ = first term = 2

d = common difference = 6 - 2 = 4

Using explicit formula:

[tex]a_{n} = a_{1} + d (n-1)[/tex]

[tex]a_{n} = 10 + 5(n-1)[/tex]

[tex]a_{n} = 10 + 5n-5[/tex]

[tex]a_{n} = 5n+ 5[/tex]

Keywords: recursive formula, explicit formula, sequence

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Final answer :

The recursive rule for the sequence is Sn =  Sn- 1 4 with S1 =  2, and the  unequivocal rule is Sn =  4n- 2. Both rules help in determining the sequence's terms either by using the  former term or directly calculating any term's value.  

Explanation:

To determine the recursive rule and the  unequivocal rule for the sequence 2, 6, 10, 14, we need to observe the pattern of the sequence and  produce formulas that can  induce the terms of the sequence. This sequence is  computation because the difference between each term is constant.

Recursive Rule

A recursive rule for a sequence defines each term grounded on the  former term. For the given sequence, each term after the first is  attained by adding 4 to the  former term. The recursive rule can be written as  Sn =  Sn- 1 4, for n> 1  where S1 =  2 is the first term.  

unequivocal Rule

An  unequivocal rule, also known as a unrestricted- form expression, defines the  utmost term of a sequence as a function of n without  demanding the  former term. For this  computation sequence, the  unequivocal rule can be determined by using the formula Sn =  a( n- 1) d, where' a' is the first term and'd' is the common difference. The  unequivocal rule for our sequence is  Sn =  2( n- 1) × 4 =  4n- 2.

An automobile dealer wants to see if there is a relationship between monthly sales and the interest rate. A random sample of 4 months was taken. The results of the sample are presented below. Monthly Sales (Y) Interest Rate (In Percent) (X) 22 9.2 20 7.6 10 10.4 45 5.3 a) Use the method of least squares to compute an estimated regression line. b) Obtain a measure of how well the estimated regression line fits the data.

Answers

Answer:

a) [tex]y=-6.254 x +75.064[/tex]  

b) r =-0.932

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]-0.932^2 =0.8687[/tex], so then the % of variation explained by the linear model is 86.87%.

Step-by-step explanation:

Assuming the following dataset:

Monthly Sales (Y)     Interest Rate (X)

       22                               9.2

       20                               7.6

       10                                10.4

       45                                5.3

Part a

And we want a linear model on this way y=mx+b, where m represent the slope and b the intercept. In order to find the slope we have this formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]  

Where:  

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]  

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]  

With these we can find the sums:  

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=278.65-\frac{32.5^2}{4}=14.5875[/tex]  

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=696.9-\frac{32.5*97}{4}=-91.225[/tex]  

And the slope would be:  

[tex]m=\frac{-91.225}{14.5875}=-6.254[/tex]  

Nowe we can find the means for x and y like this:  

[tex]\bar x= \frac{\sum x_i}{n}=\frac{32.5}{4}=8.125[/tex]  

[tex]\bar y= \frac{\sum y_i}{n}=\frac{97}{4}=24.25[/tex]  

And we can find the intercept using this:  

[tex]b=\bar y -m \bar x=24.25-(-6.254*8.125)=75.064[/tex]  

So the line would be given by:  

[tex]y=-6.254 x +75.064[/tex]  

Part b

For this case we need to calculate the correlation coefficient given by:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

[tex]r=\frac{4(696.9)-(32.5)(97)}{\sqrt{[4(278.65) -(32.5)^2][4(3009) -(97)^2]}}=-0.937[/tex]  

So then the correlation coefficient would be r =-0.932

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]-0.932^2 =0.8687[/tex], so then the % of variation explained by the linear model is 86.87%.

Use an Addition or Subtraction Formula to find the exact value of the expression, as demonstrated in Example 1. cos 41π 12.

Answers

Answer: [tex]\frac{\sqrt{6}-\sqrt{2}}{2} [/tex]

Step-by-step explanation:

We apply the formula [tex]\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y) [/tex].

Note that  [tex]\cos(\frac{41}{12}\pi)=\cos((\frac{36}{12}+\frac{7}{12})\pi)=\cos(3\pi + \frac{7}{12})\pi) [/tex]. Take  [tex]x=3\pi[/tex] and [tex]y=\frac{7}{12}\pi[/tex] in the formula above to get

[tex]\cos(\frac{41}{12}\pi)=\cos(3\pi)\cos(\frac{7}{12}\pi)-\sin(3\pi)\sin(\frac{7}{12}\pi)=(-1)\cdot \cos(\frac{7}{12}\pi)-0\cdot\sin(\frac{7}{12}\pi)=-\cos(\frac{7}{12}\pi)[/tex]

Then the value of this expression is [tex]-\cos(\frac{7}{12}\pi) [/tex]

We can use the cosine addition formula again to simplify further. Decompose the fraction in the argument as:

[tex]\cos(\frac{7}{12}\pi)=\cos((\frac{3}{12}+\frac{4}{12})\pi)=\cos((\frac{1}{4}\pi + \frac{1}{3})\pi) [/tex]

Applying the formula with [tex]x=\frac{1}{4}\pi[/tex] and [tex]y=\frac{1}{3}\pi[/tex] we obtain

[tex]\cos(\frac{7}{12}\pi)=\cos(\frac{1}{4}\pi)\cos(\frac{1}{3}\pi)-\sin(\frac{1}{4}\pi)\sin(\frac{1}{3}\pi)=\frac{\sqrt{2}}{2}\cdot\frac{1}{2} -\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{2} [/tex]

We conclude that this expression has the value [tex]-\frac{\sqrt{2}-\sqrt{6}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2} [/tex]

Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the probability that a randomly selected woman has a foot length less than 10.0 in.b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Answers

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 9.6, \sigma = 0.5[/tex].

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when [tex]X = 10[/tex].

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{10 - 9.6}{0.5}[/tex]

[tex]Z = 0.8[/tex]

[tex]Z = 0.8[/tex] has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{8 - 9.6}{0.5}[/tex]

[tex]Z = -3.2[/tex]

[tex]Z = -3.2[/tex] has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have [tex]n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1[/tex].

This probability is 1 subtracted by the pvalue of Z when [tex]X = 9.8[/tex]. So:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{9.8 - 9.6}{0.1}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the differential equation dPdt=cln(KP)P where c is a constant and K is the carrying capacity.(a) Solve this differential equation for c=0.1, K=2000, and initial population P0=500. P(t)= .(b) Compute the limiting value of the size of the population. limt→[infinity]P(t)= .(c) At what value of P does P grow fastest? P= .

Answers

Answer:

A) [tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]

B) P(t→∞)=2000

C) [tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]

Step-by-step explanation:

Given differential eq is

                      [tex]\frac{dP}{dt}=c ln (\frac{K}{P})P[/tex] --- (1)

Eq is separable

                     [tex]\frac{1}{ln (\frac{K}{P})P}dP=cdt[/tex] --- (2)

                     [tex]let \\u = ln\frac{K}{P}\\du= \frac{1}{\frac{K}{P}}(\frac{-K}{P^{2}}).dP\\du=\frac{-1}{P}.dP\\dP=-P.du[/tex]

substituting in  (2)

[tex]-\frac{du}{u}=dt[/tex]

Integrating both sides

[tex]\int {-\frac{1}{u}} \, du=\int{c}\,dt\\-ln|u|=ct +B\\ln|u|=-ct -B\\[/tex]

Back substituting value of u

[tex]ln |ln\frac{K}{P}|=-ct-B\\|ln\frac{K}{P}|=e^{-ct-B}\\ln|\frac{K}{P}|=be^{-ct}\\[/tex]---(3)

at t =0

[tex]ln|\frac{K}{P}|=be^{-ct}\\b=ln|\frac{K}{P}|\\b=ln\frac{2000}{500}\\b=ln|4|[/tex]

from (3)

[tex]ln|\frac{K}{P}|=be^{-ct}\\\frac{K}{P}=e^{ln4e^{-ct}}\\P(t)=\frac{K}{e^{ln4e^{-ct}}}[/tex]

[tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]

B) [tex]\lim{t \to \infty}[/tex]

[tex]P( {t \to \infty} )=\frac{2000}{e^{ln4e^{-0.1\infty}}}\\e^{-0.1\infty}=0\\\implies P( {t \to \infty} )=\frac{2000}{e^{0}}}\\\\P(\infty)=2000\\[/tex]

which is the carrying capacity.

C) To find the fastest growth rate we have to maximize [tex]\frac{dP}{dt}[/tex]

From given differential eq

[tex]\frac{dP}{dt}=cln|\frac{K}{P}|P[/tex]

so function to maximize is

[tex]f(P)=cln|\frac{K}{P}|P[/tex]

[tex]f'(P)=cln|\frac{K}{P}|+c\frac{1}{\frac{K}{P}}\frac{-K}{P^{2}}.P[/tex]

[tex]f'(P)=c[ln|\frac{K}{P}|-1][/tex]

To maximize find f'(P)=0

[tex]c[ln|\frac{K}{P}|-1]=0[/tex]

[tex]ln|\frac{K}{P}|=1[/tex]

[tex]\frac{K}{P}=e[/tex]

[tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]

Final answer:

The Gompertz function models population growth considering the carrying capacity. solve the Gompertz differential equation for c=0.1, K=2000, P0=500. The carrying capacity (K) gives the limiting value of the population size (2000). The time at which the population growth is fastest can be calculated by taking the derivative of the population function, setting it to zero and solving for P.

Explanation:

The Gompertz function is a model of growth that was developed to model population growth, considering the factor of carrying capacity. Instead of compound exponential growth, it models the growth as slowing down as it reaches the limit of the carrying capacity. In your particular case, solving the differential equation for the variables provided (c=0.1, K=2000, P0=500) would require the integration techniques and use of logarithmic functions.

(a): The full solution for the Gompertz function involves advanced mathematics, you should expect some intricate function in the form of P(t) = ... The particulars depend on the specifics of the integration process.

(b): The limiting value of the population size as t→infinity (limt→[infinity]P(t) will be K, which in this instance equals 2000. This is due to the concept of carrying capacity. Beyond this value, the environment/conditions can no longer support additional growth.

(c): Finding the time at which population growth is fastest involves setting the derivative of the population function to zero and solving for P. The solution P=... can be calculated using standard techniques of calculus.

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Design specifications for filling a bottled soda claim that bottles should contain 350-360 milliliters of liquid. Sample data indicate that the bottles contain an average of 355 milliliters of liquid, with a standard deviation of 2 milliliters. Is the filling operation capable of meeting the design specifications?

Answers

Answer:

It is high likely that the filling operation is capale of meeting design specifications.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 355, \sigma = 2[/tex]

Is the filling operation capable of meeting the design specifications?

It will be capable if it is highly likely that the specifications will be met. A probability is said to be high likely when it is of at least 95%.

In this case, the probability of containing between 350 and 360 ml of liquid is the pvalue of Z when X = 360 subtracted by the pvalue of Z when X = 350.

X = 360

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{360 - 355}{2}[/tex]

[tex]Z = 2.5[/tex]

[tex]Z = 2.5[/tex] has a pvalue of 0.9938.

X = 350

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{350 - 355}{2}[/tex]

[tex]Z = -2.5[/tex]

[tex]Z = -2.5[/tex] has a pvalue of 0.0062.

This means that there is a 0.9938 - 0.0062 = 0.9876 = 98.76% probability that the filling operation is capable of meeting the design specifications. It is high likely that the filling operation is capale of meeting design specifications.

Please help. I do not understand how to solve this

Answers

The answer is 2, it’s asking how are they equal to each other and how I did it I plug in the numbers for x and see if they are both the same

Answer:

X=2

Step-by-step explain

4\div 10(12-3x)=3\div 10(12x-16)

when we do cross multiply then we get

40(12-3x)=30(12x-16)

we can divide by 10 both side and solve then we get

48-12x=36x-48

and after solving this equation we get

x=2

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