Given an actual demand this period of 61, a forecast value for this period of 58, and an alpha of 0.3, what is the experimential smoothing forecast for the next period?

Answer:a is 40 degrees

b is 140 degrees

Step-by-step explanation:

The given polygon has 5 irregular sides. This means that it is an irregular pentagon. The sum of the exterior angles of a polygon is 360 degrees

The exterior angles of the given pentagon are a, 75, 65, 60 and 120. Therefore

a + 75 + 65 + 60 + 120 = 360

a + 320 = 360

Subtracting 320 from both sides of the equation, it becomes

a = 360 - 320 = 40 degrees

The sum of angles on a straight line is 180/degrees. Therefore,

a + b = 180

b = 180 - a = 180 - 40 = 140 degrees

Answer:

Her usual driving speed is 38 miles per hour.

Step-by-step explanation:

We know that:

[tex]s = \frac{d}{t}[/tex]

In which s is the speed, in miles per hour, d is the distance, in miles, and t is the time, in hours.

We have that:

At speed s, she takes two hours to drive. So

[tex]s = \frac{d}{2}[/tex]

[tex]d = 2s[/tex]

However, on one particular trip, after 40% of the drive, she had to reduce her speed by 30 miles per hour, driving at this slower speed for the rest of the trip. This particular trip took her 228 minutes.

228 minutes is 3.8 hours. So

[tex]0.4s + 0.6(s - 30) = \frac{d}{3.8}[/tex]

So

[tex]3.8(0.4s + 0.6s - 18) = d[/tex]

[tex]3.8s - 68.4 = 2s[/tex]

[tex]1.8s = 68.4[/tex]

[tex]s = 38[/tex]

Her usual driving speed is 38 miles per hour.

Answer:

[tex]s_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} }[/tex]

Step-by-step explanation:

The formula for the standard error of the difference between the estimates od the population proportions is:

[tex]s_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} }[/tex]

This is expected, as the variance of a sum (or a substraction) of two random variables is equal to the sum of the variance of the two variables.

Then, the standard error (or standard deviation) is the square root of this variance.

Answer:

D. All of the above

Step-by-step explanation:

By the given data, some persons were in the experiment and each of them was given with three meals, Low Calorie, Moderate Calorie and High Calorie. They results for the meal they liked is as below:

Low Calorie Meal got likes = 6

Moderate Calorie Meal got likes = 7

High Calorie Meal got likes = 17

so,

Option A is correct as High Calorie meal got 17 likes while Low Calorie meal got 6 likes.

Option B is also correct as Low Calorie meal got 6 likes while Moderate Calorie meal got 7 likes.

Option C is correct too as High Calorie meal got largest number of likes even more than the double of Low calorie and Moderate calorie meal so it was more than expected.

Final answer:

Based on the data provided, without performing an actual chi-squared test due to a lack of p-value or statistical data, the most supported conclusion is that participants liked the high calorie meal more than was expected (option C). This is inferred from the observed preference for the high calorie meal, which was chosen significantly more often (17 times) compared to its expected frequency (10 times).

Explanation:

The question presents a scenario where participants are asked to taste three types of meals with different calorie counts and choose the one they like best. The observed frequencies (fo) for each type of meal (low calorie, moderate calorie, and high calorie) are given as 6, 7, and 17, respectively. The expected frequencies (fe) are all 10, assuming no preference among the meals. Using a significance level of .05, we must employ a chi-squared test to determine if the observed preferences significantly differ from what was expected given no effect of calorie count.

The chi-squared test result for this would involve calculating the sum of squared differences between observed and expected frequencies, divided by the expected frequencies for each category and summing those values. However, since the data on the actual chi-squared statistic or the p-value are not provided, we cannot perform the exact calculation. Nonetheless, we can make a conclusion based on the observed and expected frequencies:

There is a notable preference for the high calorie meal, as it was chosen substantially more often than expected (17 vs. 10).

The low calorie meal was chosen less often than expected (6 vs. 10).

The moderate calorie meal was also chosen less than expected, but to a lesser extent (7 vs. 10).

Without a specific p-value or chi-squared statistic, we can't formally conclude that the participants liked the high calorie meal significantly more than the other meals, but the observed data suggest that might be the case. Also, we cannot definitively reject the possibility that the preference for the high-calorie meal is due to chance.

Therefore, the most reasonable conclusion based on the given information without formal test statistics would be option C: Participants liked the high calorie meal more than was expected.

This is in line with the general principle that a higher calorie meal tends to be preferred potentially due to higher fat content, which has been associated with better taste and satisfaction, particularly in fast-food meals. At a .05 level of significance and without the actual test statistic, option C is the most supported by the provided data.

Final answer:

The cutoff time to be in the fastest 10% of job applicants is approximately 118.0 seconds or less, calculated using a z-score of the 10th percentile in a normal distribution.

Explanation:

The student is asking about finding the cutoff time for the fastest 10% of job applicants performing a certain task, given that the time taken is normally distributed with a mean of 150 seconds and a standard deviation of 25 seconds. To solve this problem, we will use the z-score corresponding to the fastest 10% (the 10th percentile) in a normal distribution.

First, we look up the z-score for the 10th percentile in the z-table, which is approximately -1.28. Then we use the z-score formula:

z = (X - μ) / σ

Plugging in the known values:

-1.28 = (X - 150) / 25

Now, we solve for X:

X = -1.28 × 25 + 150

X = -32 + 150

X = 118 seconds

So, job applicants must complete the task in approximately 118.0 seconds or less to qualify for the advanced training.

Answer:

The answer is b): Yes, and it is highly probable.

Step-by-step Explanation:

Yes, it is highly probable that the mean BAC of all drivers involved in fatal accidents and found to have positive BAC values, is less than the legal intoxication level because blood alcohol concentration (BAC) level is not the most/only factor that determines fatal accidents. Other types of human errors are the main causes of fatal accidents; they include: side distractions, avoiding the use of helmets and seat belts, over-speeding, beating traffic red lights, overtaking in an inappropriate manner, using the wrong lane, etc. Most times, these errors are not caused by BAC levels.

Answer:

Step-by-step explanation:

Answer:

Critical value: [tex]z= 2.575[/tex]

99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).

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{1-0.99}{2} = 0.005[/tex]

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

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex] is the critical value

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}} = 2.575*\frac{0.0142}{\sqrt{17}} = 0.0089[/tex]

The lower end of the interval is the mean subtracted by M. So it is 0.704 - 0.0089 = 0.695 cc/cubic meter.

The upper end of the interval is the mean added to M. So it is 0.704 + 0.0089 = 0.713 cc/cubic meter.

So

99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).

Answer:

The standard deviation of the distribution is 492.847

Step-by-step explanation:

Consider the provided information.

A city has 1,091,953 residents. A recent census showed that 364,656 of these residents regularly use the city's public transportation system.

Therefore the value of n is 1,091,953

The probability of success is: [tex]\frac{364,656 }{1,091,953}=0.33395[/tex]

Calculate the standard deviation using the formula: [tex]\sigma=\sqrt{np(1-p)}[/tex]

Substitute the respective values.

[tex]\sigma=\sqrt{1091953(0.334)(1-0.334)}[/tex]

[tex]\sigma=\sqrt{242898.3931}[/tex]

[tex]\sigma=492.847[/tex]

Hence, the standard deviation of the distribution is 492.847

Answer:

Based on the model, the length of the wall is [tex]\frac{9}{8}[/tex] ft, the width of the wall is [tex]\frac{1}{2}[/tex] ft,  and the height of the wall is [tex]\frac{11}{8}[/tex] ft. The volume of the portion of security wall that  Tim has constructed so far is [tex]\frac{99}{128}[/tex]  cu ft.

Step-by-step explanation:

Given:

The figure constructed shows a rectangular prism made up of small wooden cubes of length [tex]\frac{1}{8}\ ft[/tex].

Width of the prism = [tex]\frac{1}{2}\ ft[/tex].

To find length , height and volume of the figure.

Solution:

From the figure we can conclude that :

Length side of the prism counts 9 cubes.

Thus, length of the prism will be given as :

⇒ [tex]\textrm{Length of each cube}\times \textrm{Number of cubes}[/tex]

⇒ [tex]\frac{1}{8}\ ft\times 9[/tex]

⇒ [tex]\frac{9}{8}\ ft[/tex]  (Answer)

Height side of the prism counts 11 cubes.

Thus, height of the prism will be given as :

⇒ [tex]\textrm{Length of each cube}\times \textrm{Number of cubes}[/tex]

⇒ [tex]\frac{1}{8}\ ft\times 11[/tex]

⇒ [tex]\frac{11}{8}\ ft[/tex] (Answer)

Volume of the prism can be given as :

⇒ [tex]Length\times width\times height[/tex]

⇒ [tex]\frac{9}{8}\ ft\times \frac{1}{2}\ ft\times \frac{11}{8}\ ft  [/tex]

⇒ [tex]\frac{99}{128}\ ft^3[/tex] (Answer)

The value of the function g(x) at x = 2.9 and x = 3.1 will be -6.57 and -3.37, respectively. And the estimation is too small in part (a).

Integration is a way of finding the total by adding or summing the components. It's a reversal of differentiation, in which we break down functions into pieces. This approach is used to calculate the total on a large scale.

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Integrate the function, then we have

∫g'(x) = ∫(x² + 7) dx

g(x) = x³/3 + 7x + c

g(3) = 3³ / 3 + 7 (3) + c

- 5 = 9 + 21 + c

c = - 30 - 5

c = -35

Then the function is given as,

g(x) = x³/3 + 7x - 35

At x = 2.9, we have

g(2.9) = (2.9)³/3 + 7(2.9) - 35

g(2.9) = -6.57

At x = 3.1, we have

g(3.1) = (3.1)³/3 + 7(3.1) - 35

g(3.1) = -3.37

The value of the function g(x) at x = 2.9 and x = 3.1 will be -6.57 and -3.37, respectively. And the estimation is too small in part (a).

More about the integration link is given below.

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Answer:

a=2

b=6

Step-by-step explanation:

Assuming a uniform distribution and that a ≤ x ≤ b, if f(x) = 1/4, then:

[tex]f(x) =\frac{1}{4}=\frac{1}{b-a}\\b-a = 4[/tex]

If P(X > 4) = 1/2, then:

[tex]\frac{1}{2} = 1-(\frac{x-a}{b-a})=1-(\frac{4-a}{4})\\a-4+4 =\frac{1}{2}*4\\ a=2\\b=4+a\\b=6[/tex]

The values for a and b are, respectively, 2 and 6.

Answer:

a) [tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]

b) [tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]

c) [tex]P(\bar X <2.52) = P(Z<-4.286)=0.000[/tex]

d) For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.

If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.

Step-by-step explanation:

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]

Let X the random variable that represent the amount of water in a bottle of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(2.55,0.035)[/tex]

a. What is the probability that an individual bottle contains less than 2.52 liters?

We are interested on this probability

[tex]P(X<2.52)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<2.52)=P(\frac{X-\mu}{\sigma}<\frac{2.52-\mu}{\sigma})[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]

b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded)

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

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

On this case  [tex]\bar X \sim N(2.55,\frac{0.035}{\sqrt{4}})[/tex]

The z score on this case is given by this formula:

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

And if we replace the values that we have we got:

[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{4}}}=-1.714[/tex]

For this case we can use a table or excel to find the probability required:

[tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]

c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.)

The z score on this case is given by this formula:

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

And if we replace the values that we have we got:

[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{25}}}=-4.286[/tex]

For this case we can use a table or excel to find the probability required:

[tex]P(\bar X <2.52) = P(Z<-4.286)=0.0000091[/tex]

d. Explain the difference in the results of (a) and (c)

For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.

If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.  

The problem involves using z-scores, normal distribution, central limit theorem and law of large numbers for probability calculations related to the amount of water in sampled bottles. The increase in sample size from 1 to 25 should show a higher probability for the sample's mean to be close to the population mean.

In this problem, we will apply the concept of the normal distribution and the central limit theorem to calculate probabilities and compare results. First, we will calculate the z-scores for (a), (b), and (c). The z-score formula is Z = (X - μ) / σ. In (a), X is 2.52 liters, μ (mean) is 2.55 liters, and σ (standard deviation) is 0.035 liters.

In (b) and (c), the σ of a sample mean is σ/√n, where n is the sample size (4 in (b) and 25 in (c)).

Next, we look up the z-score in the z-score table (or use a normal distribution calculator) to get the probabilities.

To answer part (d), probabilistically, as sample size increases, the sample mean tends to get closer to the population mean according to the law of large numbers. Hence, the probabilities in (c) should be higher than (a).

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Answer: C. 461

Step-by-step explanation:

We know that the formula to find the sample size is given by :-

[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]

, where [tex]\sigma[/tex] = Population standard deviation from prior study.

E = margin of error.

z* = Critical value.

As per given , we have

[tex]\sigma=125[/tex]

E= 15

Significance level : [tex]\alpha=0.01[/tex]

Critical value (Two tailed)=[tex]z^*=z_{\alpha/2}=z_{0.005}=2.576[/tex]

Now , Required sample size = [tex]n=(\dfrac{(2.576)\cdot 125}{15})^2[/tex]

[tex]n=(21.4666666667)^2\\\\ n=460.817777779\approx461[/tex] [Round to next integer]

Hence, the required sample size to be taken is 461.

Correct answer = C. 461

Answer:

$6

Explanation:

The standard deviation gives an idea of the dispertion of values in those 12 month. A hight value of standard deviation means that the prices changed in big increments ( for example one month is $20, other month is $60 and other $85). If that  would be the case, is important to investors to know it and the deviation should be reported.

So, given that it isnt reported, we can say that the price vary but only a few dollars, resulting in a small standard deviation such as $6.

Final answer:

A standard deviation of $6 for a blue chip stock is more likely than $26 or $60, as blue chip stocks are typically stable with less price volatility.

Explanation:

Regarding the standard deviation of the blue chip stock price, it is unlikely that the standard deviation would be very high considering the nature of blue chip stocks. Blue chip stocks are known for their stability and are typically less volatile than other types of stocks. Therefore, a standard deviation of $60 would suggest a very high level of volatility, which is uncharacteristic of blue chip stocks. A standard deviation of $6 is more plausible because it indicates minor fluctuations around the average price of $72, which is more in line with the expected behavior of blue chip stocks. A standard deviation of $26 is possible but less likely than $6, given that such a value still represents a fairly high level of volatility for blue chip stocks, which are commonly considered as safe, long-term investments with relatively steady prices.

Answer:

Null hypothesis:[tex]\mu \geq 4.0[/tex]    

Alternative hypothesis:[tex]\mu < 4.00[/tex]    

[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]    

[tex]df=n-1=45-1=44[/tex]

Since is a left-sided test the p value would be:    

[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/tex]    

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.  

Step-by-step explanation:

Data given and notation    

[tex]\bar X=3.48[/tex] represent the sample mean

[tex]s=1.150075[/tex] represent the sample standard deviation    

[tex]n=45[/tex] sample size    

[tex]\mu_o =4.00[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

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

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 4.00 :    

Null hypothesis:[tex]\mu \geq 4.0[/tex]    

Alternative hypothesis:[tex]\mu < 4.00[/tex]    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]    

P-value    

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=45-1=44[/tex]

Since is a left-sided test the p value would be:    

[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/tex]    

Conclusion    

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.    

The price of one ticket is $ 62.2

Given that a performer expects to sell 5000 tickets for an upcoming event

They want to make a total of $ 311, 000 in sales from these tickets

To find: price of one ticket

Let us assume that all tickets have the same price

Let "a" be the price of one ticket

So the total sales price of $ 311, 000 is obtained from product of 5000 tickets and price of one ticket

[tex]\text {total sales price }=5000 \times \text { price of one ticket }[/tex]

[tex]311000 = 5000 \times a\\\\a = \frac{311000}{5000}\\\\a = 62.2[/tex]

Thus the price of one ticket is $ 62.2

Answer:

The mean weight is not greater than 16 ounces

Step-by-step explanation:

Null hypothesis: Mean weight is greater than 16 ounces

Alternate hypothesis: Mean weight is not greater than 16 ounces.

Sample mean= 16.04 ounces

Assumed mean= 16 ounces

Significance level = 0.05

Sample standard deviation = 0.08 ounces

Number of samples= 35

To test the claim about the mean, Z = (sample mean - assumed mean) ÷ (standard deviation÷ √number of samples)

Z = (16.04 - 16) ÷ (0.08 ÷ √35)

Z = 0.04 ÷ (0.08 ÷ 5.92)

Z= 0.04 ÷ 0.014 = 2.86

Z = 2.86

It is a one-tailed test, the critical region is an area of 0.05 (the significance level). The critical value of the critical region is 1.64

Since the computed Z 2.86 is greater than the critical value 1.64, the null hypothesis is rejected.

Conclusion

The mean weight is not greater than 16 ounces

Answer:

We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 100 pounds

Sample mean, [tex]\bar{x}[/tex] = 98 pounds

Sample size, n = 35

Alpha, α = 0.05

Population standard deviation, σ = 7.5 pounds

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 100\text{ pounds}\\H_A: \mu < 100\text{ pounds}[/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{98 - 100}{\frac{7.5}{\sqrt{35}} } = -1.577[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -1.64[/tex]

We calculate the p-value with the help of standard normal table.

P-value = 0.057398

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

We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.

To calculate the p-value for the hypothesis test, we calculate the t-score using the given values and the formula. Then, we use the t-distribution table or a calculator to find the p-value associated with the t-score and degrees of freedom. The calculated p-value is approximately 0.1093.

In this problem, we are given the mean and standard deviation of a normal distribution for the weights of seventh-graders. The researcher believes that the new breakfast and lunch program has decreased the average weight. We are asked to calculate the p-value for an appropriate hypothesis test.

To calculate the p-value, we can use the t-distribution since the population standard deviation is unknown. We can calculate the t-score using the formula:

t = (sample mean - population mean) / (sample standard deviation / √sample size)

Plugging in the given values, we get t = (98 - 100) / (7.5 / √35) = -1.654. The degrees of freedom for this test is (sample size - 1) = (35 - 1) = 34.

Using the t-distribution table or a calculator, we can find the p-value associated with a t-score of -1.654 and 34 degrees of freedom. The p-value is approximately 0.1093.

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Multiply his rate by number of hours:

30.40 x 30 = $912

Shane would make $912 if he works 30 hours at his normal work rate.

To calculate how much Shane makes if he works 30 hours at an hourly wage of $30.40, we can multiply his hourly wage by the number of hours he works:

$30.40/hour x 30 hours = $912

Therefore, Shane would make $912 if he works 30 hours at his normal work rate.

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Answer:

y= -2x + 16

Step-by-step explanation:

multiple the -2 with everything in parentheses then get -2x and 6 add 10 to both sides then then bring everything down to get y= -2x + 16

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

[tex]n_1 [/tex] represent the sample after the preparation

[tex]n_2 [/tex] represent the sample without preparation  

[tex]\bar x_1 =678[/tex] represent the mean sample after preparation

[tex]\bar x_2 =1837[/tex] represent the mean sample without preparation

[tex]s_1 =197[/tex] represent the sample deviation after preparation

[tex]s_2 =328[/tex] represent the sample deviation without preparation

[tex]\alpha=0.1[/tex] represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex]

The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

[tex]SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}[/tex]  

After replace in the formula for the confidence interval we got this:

[tex]3.0 < \mu_1 -\mu_2 <13.0 [/tex]

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Answer:

[tex] y= 2x +1[/tex]

By direct comparison we can see that m =2 (slope) and b =1 (intercept)

Step-by-step explanation:

Assuming the following function:

[tex] y-2x =1[/tex]

We want to find the following general equation for a linear model:

[tex]y = mx +b [/tex]

On this case we just need to add 2x on both sides of the original equation and we got:

[tex] y= 2x +1[/tex]

By direct comparison we can see that m =2 and b =1

[tex] mx + b = 2x +1[/tex]

[tex]mx=2x , m =2[/tex]

[tex] b =1[/tex]

the value of m on this case represnt the slope and b the intercept.

The slope is defined by the following formula:

[tex] m =\frac{\Delta y}{\Delta x}[/tex]

And is the interpretation is the rate of change of y respect to x, can be positive or negative. Or the increase/decrease of y when x increase 1 unit.

And the value b=1 represent the y intercept, that means if x=0 then y =1

Answer:

A region R is revolved about the y-axis. The volume of the resulting solid could (in principle) be found using the disk/washer method and integrating with respect to y or using the shell method and integrating with respect to x.

Step-by-step explanation:

We assume this question :

Fill in the blanks: A region R is revolved about the y-axis. The volume of the resulting solid could (in principle) be found using the disk/washer method and integrating with respect to _ or using the shell method and integrating with respect to _____.

We can calculate the volume of a region when revolved about y-axis with two common methods: washer method and shell method. We need to take in count, that the general formula for these methods are different respect to the variable used to calculate the volume.

Since the region R is revolved about y-axis, disk/washer method needs to calculate the integral respect to y, and by the other hand the shell method will calculate the integral respect to x.

Washer method

[tex]V\approx \sum_{k=1}^n \pi r^2 h = \sum_{k=1}^n \pi f(y_k)^2 dy[/tex]

[tex]V= \pi \int_{c}^d f(y)^2 dy[/tex]

Shell method

[tex]V = \lim_{n\to\infty} 2\pi x_i f(x_i)\Delta x =\int_{a}^b 2\pi x f(x) dx[/tex]

So then the correct answer would be:

A region R is revolved about the y-axis. The volume of the resulting solid could (in principle) be found using the disk/washer method and integrating with respect to y or using the shell method and integrating with respect to x.

Answer:

z-test statistic is 3.38

Step-by-step explanation:

Null and alternative hypotheses are:

[tex]H_{0}[/tex]: the proportion of papers that lacked statistical assistance  sent back without review is the same as the proportion of papers with statistical help sent back without review

[tex]H_{a}[/tex]: the proportion of papers that lacked statistical assistance sent back without review is different than the proportion of papers with statistical help sent back without review

Test statistic can be found using the equation:

[tex]z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

Then [tex]z=\frac{0.71-0.57}{\sqrt{{0.61*0.39*(\frac{1}{190} +\frac{1}{514}) }}}[/tex] ≈ 3.38

Answer: At-least 88.89%

Step-by-step explanation:

As per given , we have

Population mean : [tex]\mu=\$3.73[/tex]

Standard deviation :  [tex]\sigma=\$0.10[/tex]

Now , $3.43= $⁢3.73- 3(0.10) = [tex]\mu-3\sigma[/tex]

$⁢4.03 = $⁢3.73+3(0.10) = [tex]\mu+3\sigma[/tex]

i.e. $3.43 is 3 standard deviations below mean and $⁢4.03 is 3 standard deviations above mean .

To find :  the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03.

i.e. to find minimum percentage of stores that sell a gallon of milk  lies within 3 standard deviations from mean.

According to Chebyshev, At-least [tex](1-\dfrac{1}{k^2})[/tex] of the values  lies with in [tex]k\sigma[/tex] from mean.

For k= 3

At-least [tex](1-\dfrac{1}{3^2})[/tex] of the values lies within [tex]3\sigma[/tex] from mean.

[tex]1-\dfrac{1}{3^2}=1-\dfrac{1}{9}=\dfrac{8}{9}[/tex]

In percent = [tex]\dfrac{8}{9}\times100\%\approx88.89\%[/tex]

Hence, the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03 =  At-least  88.89%

Final answer:

Using Chebyshev's Theorem, the minimum percentage of stores selling a gallon of milk between $3.43 and $4.03 is at least 88.9% when the mean price is $3.73 and the standard deviation is $0.10.

Explanation:

Using Chebyshev's Theorem, we can determine the minimum percentage of stores selling a gallon of milk within a certain range of prices given the mean and standard deviation. The theorem states that for any number k, where k > 1, at least (1 - 1/k²) of the data values will fall within k standard deviations of the mean. In this case, the range of prices is from $3.43 to $4.03, which is $0.30 away from the mean of $3.73 on either side.

To find k, we divide the distance from the mean by the standard deviation: k = 0.30 / 0.10 = 3. Thus, at least (1 - 1/3²) or (1 - 1/9) of the stores sell a gallon of milk within this range. Calculating this, we get at least (1 - 1/9) = 8/9 or approximately 88.9% of stores.

Therefore, according to Chebyshev's Theorem, the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03 is at least 88.9%.

Answer: a. 4

Step-by-step explanation:

The number of degrees of freedom represents the number of data-values in the evaluation of a test-statistic that are independent to vary.

The degree of freedom for chi-square test = n-1 , n= Sample size.

Given : A study was conducted to examine whether the proportion of females was the same for five groups (Groups A, B, C, D, and E).

i.e. n= 5

Then, the number of degrees of freedom the χ2 test statistic have when testing the hypothesis that the proportions in each group are all equal = 5-1=4

Hence, the correct answer is OPTION a. 4 .

Answer:

V = π (rh² − ⅓h³)

Step-by-step explanation:

Draw a cross section of the bowl.  Cut a thin, horizontal slice of the water.  This slice is a circular disc of radius x and thickness dy.  It is position a distance of y from the bottom of the bowl.  The volume of this slice is:

dV = πx² dy

By drawing a right triangle, we can define x in terms of y:

x² + (r−y)² = r²

x² + r² − 2ry + y² = r²

x² = 2ry − y²

Substitute:

dV = π (2ry − y²) dy

The total volume of the water is the sum of all the slices from y=0 to y=h.

V = ∫₀ʰ π (2ry − y²) dy

V = π ∫₀ʰ (2ry − y²) dy

V = π (ry² − ⅓y³) |₀ʰ

V = π (rh² − ⅓h³)

Final answer:

The formula to measure the volume of water in a hemispherical bowl is V = (2/3)πr²h.

Explanation:

The formula to measure the volume of water in a hemispherical bowl is V = (2/3)πr2h, where V is the volume, π is a constant approximately equal to 3.14159, r is the radius of the bowl, and h is the depth of the water.

1. The volume of a half of the globe is given by V = 2/3 * π * r³. However, since we are only interested in measuring the water's volume—which represents a portion of the hemisphere—we will need to modify the formula accordingly.

2. Within the larger hemisphere, a smaller one is created by the water's depth, h. To ascertain the volume of this more modest half of the globe, we can take away it from the volume of the bigger side of the equator.

3. The span of the more modest half of the globe is likewise r, since it has a similar ebb and flow as the bigger side of the equator.

4. The level of the more modest side of the equator, h, is equivalent to the profundity of the water.

5. To find the volume of the more modest side of the equator, we utilize the recipe V = 2/3 * π * r³ and substitute the sweep r and level h.

6. The volume of the water is around 50% of the volume of the more modest half of the globe since it just fills one side of the bowl.

7. In this way, the volume of the water in the hemispherical bowl is given by 1/2 * (2/3 * π * r³) = 1/3 * π * r³.

8. To represent the profundity of the water, we increase the volume by the proportion of the level h to the sweep r, giving us 1/3 * π * r² * h.

9. We can further simplify by rewriting the formula as 1/2 *  * h2 * (3r - h).

Answer:

Step-by-step explanation:

1) in a rhombus, the diagonals bisect each other at the midpoint, forming 4 right angles.

MK = NK + NM

Since MK = 24 and NK = NM, then,

NK = 24/2 = 12

JL = NJ + NL

Since JL = 20 and NJ = NMlL, then,

NL = 20/2 = 10

Looking at triangle LMN, it is a right angle triangle. Applying Pythagoras theorem,

ML^2 = 10^2 + 12^2 = 100 + 144 = 244

ML = √244 = 15.62

Since all the sides of the rhombus are equal, then

MJ = ML = 15.62

Angle KNL = angle JML = 90 degrees. This is so because the diagonals are perpendicular)

Angle KJL = 90 - angle MJL

Angle KJL = 90 - 50 = 40 degrees.

Since the opposite angles of the rhombus are equal,

Angle MLK = angle angle MJK = 50 + 40 = 90 degrees

2) since all the sides of a rhombus are equal,

5x + 16 = 9x - 32

9x - 5x = 32 + 16

4x = 48

x = 48/4 = 12

Since PQ = NR = 5x + 16, then

PQ = 5×12 + 16 = 60 + 16

PQ = 76

3) Triangle XYZ is an isosceles triangle. This means that its base angles, XZY and ZXY are equal. The sum of angles in a triangle is 180 degrees. Therefore,

Angle XZY + angle ZXY = 180 - 136

Angle XZY + angle ZXY = 44

Angle XZY = 44/2 = 22

Therefore,

10x - 8 = 22

10x = 22 + 8 = 30

x = 30/10 = 3

Answer:

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]

[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/tex]

So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between Drug A and Drug B is equal to 0.

Step-by-step explanation:

Data given

              1   2   3    4    5   6   7

Drug A   6   3   4   5    7    1    4

Drug B   5    1   5   5    5   2    2

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value for A , y = test value for B

The system of hypothesis for this case are:

Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]

Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]

The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:

d: -1, -2, 1, 0, -2, 1, -2

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714[/tex]

The third step would be calculate the standard deviation for the differences, and we got:

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.38[/tex]

The 4 step is calculate the statistic given by :

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]

The next step is calculate the degrees of freedom given by:

[tex]df=n-1=7-1=6[/tex]

Now we can calculate the p value, since we have a two tailed test the p value is given by:

[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/tex]

So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between Drug A and Drug B is equal to 0.