A study was made comparing the cost of a one-bedroom apartment in philadelphia with the cost of similar apartments in Baltimore. A sample of 30 apartments in Philadelphia showed a sample mean of $950 with a standard devation of $50. A sample of 25 apartments in Baltimore showed a sample mean of $915 and a sample stand deviation of $45. Test to see if there is a significant difference in mean rental rate between the two cities. Use a 5% leve of significance.What is your conclusion?

Final answer:

To find the height of the car, we can rearrange the equation for potential energy and solve for h. Using the given values of 200 J for potential energy and 300 N for weight, the height of the car is approximately 0.683 meters.

Explanation:

To calculate the height of a car that weighs 300 Newtons and has 200 joules of gravitational potential energy, we can use the formula for potential energy, PE = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height. Since we know the potential energy and the weight (force) of the car, we can rearrange the formula to solve for h. Here's how:

Therefore, the height of the car is approximately 0.683 meters.

Answer:

Step-by-step explanation:

Given

Economic system Q is given by

[tex]Q=cL^aK^b[/tex]

also [tex]a+b=1[/tex]

if [tex]Q=12,200[/tex]

[tex]a=\frac{1}{6}[/tex]

[tex]b=\frac{5}{6}[/tex]

[tex]c=42[/tex]

substitute these values

[tex]12,200=42\times (L)^{\frac{1}{6}}K^{\frac{5}{6}}[/tex]

[tex](L)^{\frac{1}{6}}K^{\frac{5}{6}}=\frac{12,200}{42}[/tex]

[tex]K^{\frac{5}{6}}=\frac{12,200}{42(L)^{\frac{1}{6}}}[/tex]

[tex]K=(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^{5}}[/tex]

differentiate w.r.t to L to get [tex]\frac{dK}{dL}[/tex]

[tex]\frac{dK}{dL}=(\frac{12,200}{42})^{\frac{6}{5}}\times (-5)\times L^{-6}[/tex]

[tex]\frac{dK}{dL}=-5(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^6}[/tex]

[tex]\frac{dK}{dL}=-\frac{4515.466}{L^6}[/tex]

Answer:

Part (a): The mean of the distribution is 2.

Part (b): The standard deviation of the distribution is 1.525

Step-by-step explanation:

Consider the provided information.

Part (a) What is the mean of the distribution?

The mean of the sampling distribution of [tex](\bar x_1 -\bar x_2)[/tex] is (µ₁ − µ₂).

It is given that μ₁ = 31, σ₁ = 3, μ₂ = 29, and σ₂ = 2.

Mean of distribution = (x₁ - x₂)

Mean of distribution = (31 - 29) = 2

Hence, the mean of the distribution is 2.

Part (b) What is the standard deviation of the distribution?

If the two samples are independent, the standard deviation of the sampling distribution is: [tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}\right)}}[/tex]

Substitute the respective values in the above formula.

[tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{3^2}{4}+\dfrac{2^2}{53}\right)}}[/tex]

[tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{9}{4}+\dfrac{4}{53}\right)}}[/tex]

[tex]\sigma_{(x_1-x_2)}\approx1.525[/tex]

Hence, the standard deviation of the distribution is 1.525

The approximate sampling distribution of x1 - x2 has a mean equal to μ1 - μ2 and a standard deviation equal to √((σ1^2/n1) + (σ2^2/n2)).

A sampling distribution is the probability distribution of a statistic, like the mean or proportion, calculated from multiple samples of the same size drawn from a population. It provides insights into the variability of the statistic and is fundamental in statistical inference.

The approximate sampling distribution of x1 - x2 has a mean equal to the difference of the population means, μ1 - μ2. The standard deviation of the distribution is equal to the square root of the sum of the variances of the two populations divided by their respective sample sizes, √((σ1^2/n1) + (σ2^2/n2)).

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Answer:

n=50

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

[tex]\bar X=16.10[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=1.8 represent the sample standard deviation

n=25 represent the sample size  

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

We need to find the degrees of freedom given by:

[tex]df=n-1=25-1=24[/tex]

Since the Confidence is 0.95 or 95%, 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: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=\pm 1.96[/tex]

Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =0.5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

Replacing into formula (b) we got:

[tex]n=(\frac{1.96(1.8)}{0.5})^2 =49.79 \approx 50[/tex]

So the answer for this case would be n=50 rounded up to the nearest integer

The value of y when x = 5 is 45

Given that y varies directly with x. This can be written mathematically as:

y  ∝  x

[tex]y = k \times x[/tex]  ---- eqn 1

Where "k" is the constant of propotionality

Given y = 27 when x = 3. Substitute these values in eqn 1

[tex]27 = k \times 3\\\\k = \frac{27}{3} = 9[/tex]

Substitute k = 9 in eqn 1

y = 9x ----- eqn 2

To find: y = ? and x = 5

Substitute x = 5 in eqn 2

[tex]y = 9 \times 5 = 45[/tex]

Thus the value of y when x = 5 is 45

Answer:

Correct answer is (a). An online research company puts out a survey asking people two questions. First, it asks whether they buy groceries online. Second, it asks whether they own a car or not. The data are collected and recorded in a contingency table. The company wants to determine if there is a relationship between buying groceries online and car ownership

Step-by-step explanation:

The chi-square distribution can be used to perform the goodness-of-fit test, Closeness of observed data to expected data of the model.

Final answer:

The expectation of the time at which the plumber completes the project is 4:30 PM, and the variance is 901.33 [tex]minutes^2[/tex]

Explanation:

To find the expectation and variance of the time at which the plumber completes the project, we need to use the properties of the uniform and exponential distributions.

1. The arrival time of the plumber is uniformly distributed between 1 PM and 7 PM, which means it has a continuous uniform distribution. The expectation of a continuous uniform distribution is calculated as the average of the lower and upper bounds, so the expectation of the arrival time is (1 + 7) / 2 = 4 PM. The variance of a continuous uniform distribution is calculated as [tex](upper bound - lower bound)^2 / 12,[/tex] so the variance of the arrival time is [tex](7 - 1)^2 / 12 = 1.33.[/tex]

2. The time it takes to fix the broken faucet is exponentially distributed with a mean of 30 minutes, which means it has an exponential distribution. The expectation of an exponential distribution is equal to its mean, so the expectation of the fixing time is 30 minutes. The variance of an exponential distribution is equal to the square of its mean, so the variance of the fixing time is [tex]30^2 = 900 minutes^2[/tex].

Since the arrival time and fixing time are independent, the total time at which the plumber completes the project is the sum of the arrival time and fixing time. The expectation of the total time is the sum of the expectations, which is 4 PM + 30 minutes = 4:30 PM. The variance of the total time is the sum of the variances, which is 1.33 + 900 = 901.33 [tex]minutes^2.[/tex]

Answer:

a) There is a 13% probability that a student has taken 2 or more semesters of Calculus.

b) 45% probability that a student has taken some calculus.

c) 87% probability that a student has taken no more than one semester of calculus.

Step-by-step explanation:

We have these following probabilities:

A 55% that a student hast never taken a Calculus course.

A 32% probability that a student has taken one semester of a Calculus course.

A 100-(55+32) = 13% probability that a student has taken 2 or more semesters of Calculus.

a) two or more semesters of Calculus?

There is a 13% probability that a student has taken 2 or more semesters of Calculus.

b) some Calculus?

At least one semester.

So there is a 32+13 = 45% probability that a student has taken some calculus.

c) no more than one semester of Calculus?

At most one semester.

So 55+32 = 87% probability that a student has taken no more than one semester of calculus.

The correct probabilities for the given scenarios are:

a) The probability that the first groupmate you meet has studied two or more semesters of Calculus is [tex]$\boxed{\frac{13}{100}}$[/tex].

 b) The probability that the first groupmate you meet has studied some Calculus (which includes those who have taken only one semester and those who have taken two or more semesters) is [tex]$\boxed{\frac{45}{100}}$[/tex].

c) The probability that the first groupmate you meet has studied no more than one semester of Calculus is [tex]$\boxed{\frac{32}{100}}$[/tex].

To calculate these probabilities, we use the information provided by the professor:

- 55% of the students have never taken a Calculus course.

- 32% have taken only one semester of Calculus.

- The rest, which is 100% - (55% + 32%) = 13%, have taken two or more semesters of Calculus.

Now, let's calculate the probabilities for each part of the question:

a) The probability that the first groupmate has studied two or more semesters of Calculus is the percentage of students who have taken two or more semesters, which is 13%. In probability terms, this is [tex]$\frac{13}{100}$[/tex]

b) The probability that the first groupmate has studied some Calculus is the sum of the percentages of students who have taken at least one semester of Calculus. This includes both those who have taken only one semester (32%) and those who have taken two or more semesters (13%). Therefore, the probability is 32% + 13% = 45%. In probability terms, this is [tex]$\frac{45}{100}$[/tex].

c) The probability that the first groupmate has studied no more than one semester of Calculus is the sum of the percentages of students who have never taken Calculus and those who have taken only one semester. This is 55% + 32% = 87%. However, since we are looking for the probability of no more than one semester, we exclude those who have taken two or more semesters, so we subtract the 13% from the 87%, giving us 87% - 13% = 74%. But this calculation is incorrect because we have double-counted the 32% of students who have taken only one semester. The correct calculation is to consider only the students who have never taken Calculus (55%) and those who have taken only one semester (32%), which gives us 55% + 32% - 0% (since we do not need to subtract any percentage this time) = 87%. In probability terms, this is [tex]$\frac{32}{100}$[/tex], as we are only considering those who have taken no Calculus or just one semester.

Answer:

0

Step-by-step explanation:

8+3=11.

11-11=0

So the answer is zero.

Hope that helps!

Answer:

P-value = 0.4324

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 8 ounces

Sample mean, [tex]\bar{x}[/tex] = 7.983 ounces

Sample size, n = 48

Population standard deviation, σ = 0.15 ounces

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 8\text{ ounces}\\H_A: \mu \neq 8\text{ ounces}[/tex]

We use Two-tailed 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{7.983 - 8}{\frac{0.15}{\sqrt{48}} } = -0.7851[/tex]

Now, we calculate the p-value with the help of standard normal z table.

P-value = 0.4324

Final answer:

To calculate the P-Value, we need to perform a hypothesis test. The test statistic is calculated by subtracting the hypothesized mean from the sample mean and dividing by the standard error. The P-Value is the probability of observing a test statistic as extreme as the calculated value or more extreme under the null hypothesis.

Explanation:

To calculate the P-Value, we need to perform a hypothesis test. The null hypothesis, H0, is that the mean amount dispensed is equal to 8 ounces, while the alternative hypothesis, Ha, is that the mean amount dispensed is different from 8 ounces.

Using the given information, we can calculate the test statistic, which is the standardized value of the sample mean. First, we calculate the standard error of the sample mean, which is the standard deviation divided by the square root of the sample size. In this case, the standard error is 0.15 / sqrt(48) = 0.021.

The test statistic is then calculated by subtracting the hypothesized mean (8 ounces) from the sample mean (7.983 ounces) and dividing by the standard error. So, the test statistic is (7.983 - 8) / 0.021 = -0.619.

To calculate the P-Value, we need to find the probability of observing a test statistic as extreme as -0.619 (or more extreme) under the null hypothesis. We can use a standard normal distribution table or a calculator to find the corresponding area under the curve. From the table or calculator, we find that the area to the left of -0.619 is 0.269.

Since the alternative hypothesis is two-sided (the mean could be either greater or less than 8 ounces), we double the area to get the P-Value. So, the P-Value is 2 * 0.269 = 0.538.

Answer:

a. 8.66 minutes

Step-by-step explanation:

Since the flight times are uniformly distributed, the standard deviation can be calculated as follows:

[tex]\sigma = \frac{b-a}{\sqrt{12}}[/tex]

Where 'b' is the maximum flight time (150 minutes) and 'a' is the minimum flight time (120 minutes):

[tex]\sigma = \frac{150-120}{\sqrt{12}}\\\sigma = 8.66\ minutes[/tex]

The distribution's standard deviation is 8.66 minutes.

Answer: 10 guests, $250

Step-by-step explanation:

Use the equation y=mx+b

Plug in what you are given.

For caterer A the cost per guest is $15. This is the m value. The b value is the additional fee for setting up the tables, $100. Your equation should be y=15x+100.

Caterer B charges $20 per guest and a $50 fee to set up tables. The equation will be y=20x+50.

To find the number of guests that will result in an equal cost, set the equations equal to one another. Solve for x.

15x+100=20x+50

Do this by getting the x’s all on one side (I subtracted 15x on both sides). Then subtract the 50 from both sides. This results in the equation 50=5x. Dividing by 5 will make x equal to 10.

To find the cost at this number of guests plug 10 into either equation y=15x+100 or y=20x+50. This results in a cost of $250.

Answer: 10 guests, $250

Step-by-step explanation:

Use the equation y=mx+b

Plug in what you are given.

For caterer A the cost per guest is $15. This is the m value. The b value is the additional fee for setting up the tables, $100. Your equation should be y=15x+100.

Caterer B charges $20 per guest and a $50 fee to set up tables. The equation will be y=20x+50.

To find the number of guests that will result in an equal cost, set the equations equal to one another. Solve for x.

15x+100=20x+50

Do this by getting the x’s all on one side (I subtracted 15x on both sides). Then subtract the 50 from both sides. This results in the equation 50=5x. Dividing by 5 will make x equal to 10.

Answer:

14

Step-by-step explanation:

Given that a package delivery company purchased 14 trucks at the same time. Five trucks were purchased from manufacturer A, four from manufacturer B, and five from manufacturer C. The cost of maintaining each truck was recorded.

No of companies  = 3

No of items in total = 14

Total df = 17-1=16

degrees of freedom between groups = 3-1 =2

Hence degrees of freedom for denominator

=[tex]16-2 =14[/tex]

Answer:

0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.5 minutes

Standard Deviation, σ = 2 minutes

Since the sample size is large, by central limit theorem, the distribution of sample means is approximately normal.

[tex]P(\sum x_{i} > 4)\\P(\sum x_i > 4\times 60\text{ minutes})\\\\P(\displaystyle\frac{1}{100}\sum x_i > \frac{4\times 60}{100}\text{ minutes})\\\\P(\bar{x} > 2.4)[/tex]

Formula:  

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

P(it will take more than 4 hours to process the orders of 100 people)  

P(x > 2.4)  

[tex]P( x > 2.4) = P( z > \displaystyle\frac{2.4-2.5}{\frac{2}{\sqrt{100}}}) = P(z > -0.5)[/tex]  

Calculation the value from standard normal z table, we have,  [tex]P(x > 2.4) = 1 - 0.3085 = 0.6915= 69.15\%[/tex]

0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.

Final answer:

Using the Central Limit Theorem, the mean and standard deviation for 100 customers were calculated. The z-score was computed to find the probability from the standard normal distribution. The probability of taking more than 240 minutes to serve 100 customers is less than 0.5.

Explanation:

To solve this problem, we need to use the Central Limit Theorem which states that the sum or the average of a large number of independent and identically distributed random variables will be approximately normally distributed, irrespective of the original distribution of the variables.

Step by Step Solution:

First, we calculate the mean total time to process orders for 100 people. Mean total time = mean time per customer * number of customers = 2.5 minutes * 100 = 250 minutes.

Next, we calculate the standard deviation of the total time for 100 customers. Since the customer service times are independent, we can use the formula for the standard deviation of the sum of independent random variables: Standard deviation of the total time = standard deviation per customer * sqrt(number of customers) = 2 minutes * sqrt(100) = 20 minutes.

To find out if it will take more than 4 hours (which is 240 minutes), we find the z-score: z = (Total minutes to consider - Mean total time) / Standard deviation of total time = (240 - 250) / 20 = -0.5.

The probability of service taking more than 4 hours is the same as the probability of the sum being greater than 240 minutes. We use the z-score to find this probability from the standard normal distribution table or using a calculator with normal distribution functions.

Since a z-score of -0.5 corresponds to a probability higher than 0.5 (due to the symmetry of the normal distribution), the probability of taking less than 240 minutes is above 0.5. Therefore, the probability of taking more than 240 minutes is less than 0.5.

Answer:

a)0.1714 b)0.504 c)0.5 d)0.2

Step-by-step explanation:

Total numbers between 1000 and 9999 (both inclusive)= 9000.

Probability=[tex]\frac{No.OfFavourableOutcomes}{TotalNo.OfOutcomes}[/tex]

a) Divisible by 5 and not by 7 means, that the number is divisible by 5 but not by 35.

Total numbers between 1000 and 9999 divisible by 5= 1800

Total numbers between 1000 and 9999 divisible by 35=257

Total numbers between 1000 and 9999 divisible by 5 but not by 35=         1800 - 257= 1543

Probability=[tex]\frac{1543}{9000}[/tex]=0.1714

b) Finding the number of numbers with distinct digits,

We can find this by the method that how many digits can be placed in each place,

In the thousands place=9( 0 cannot be placed)

In the hundreds place=9( The digit inserted in the thousands place, cannot be put but 0 can be put now)

In the tens place=8( Numbers placed in hundreds and thousands place cannot be put)

In the ones place=7(The rest 3 numbers can't be put)

Total Favorable Outcomes=9 x 9 x 8 x 7 =4536

Probability=[tex]\frac{4536}{9000}[/tex]=[tex]\frac{63}{125}[/tex]=0.504

c) Total number of even numbers between 1000 and 9999= 4500

Probability=[tex]\frac{4500}{9000}[/tex]=[tex]\frac{1}{2}[/tex]=0.5

d) Total numbers between 1000 and 9999 divisible by 5= 1800

Probability= [tex]\frac{1800}{9000}[/tex]=[tex]\frac{1}{5}[/tex]=0.2

Based on the given data,  cannot conclude that the financial advisor's belief is accurate. Since the test statistic (Z-score) of 0.4483 is less than the critical value of 1.645, fail to reject the null hypothesis

To formulate a one-sample hypothesis test for a proportion, we can use the following steps:

Step 1: State the null hypothesis (H0) and the alternative hypothesis (H1).

Step 2: Determine the significance level (α) for the hypothesis test.

The significance level represents the probability of rejecting the null hypothesis when it is true.

Let's assume a significance level of α = 0.05, which is a commonly used value.

Step 3: Collect the data and calculate the test statistic.

[tex]\bar{p}= 21 / 33[/tex]

On dividing gives:

[tex]\bar{p}\approx 0.6364[/tex]

Standard Error (SE) [tex]= \sqrt{(\bar{p} * (1 -\bar{p})) / n}[/tex]

plugging given data gives:

[tex]= \sqrt{(0.6364 * (1 - 0.6364)) / 33}[/tex]

On simplifying gives:

≈ 0.0811

Step 4: Determine the critical value or the p-value.

To determine the critical value,  to use the Z-table . For a significance level of α = 0.05 (one-tailed),

the critical value corresponds to a z-score of approximately 1.645.

Step 5: Make a decision and interpret the results.

Calculate the test statistic (Z-score):

[tex]Z = (\bar{p} - p) / SE[/tex]

Plugging the given values gives:

[tex]= (0.6364 - 0.6) / 0.0811[/tex]

On simplifying gives:

[tex]\approx 0.4483[/tex]

Since the test statistic (Z-score) of 0.4483 is less than the critical value of 1.645, fail to reject the null hypothesis. There is not enough evidence to support the claim that the proportion of risk-averse investors is greater than 0.6 at a significance level of 0.05.

Therefore, based on the given data, we cannot conclude that the financial advisor's belief is accurate.

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Answer:

[tex]p_v =2*P(t_{(8)}<-1.195)=0.266[/tex]

c. should not be rejected

Step-by-step explanation:

1) Data given and notation

The data given is:

Method 1 : 7,5,6,7,5

Method 2: 5,9,8,7,6

[tex]\bar X_{1}=6[/tex] represent the mean for the method 1

[tex]\bar X_{2}=7[/tex] represent the mean for the method 2

[tex]s_{1}=1[/tex] represent the sample standard deviation for the method 1

[tex]s_{2}=1.58[/tex] represent the sample standard deviation for the method 2

[tex]n_{1}=5[/tex] sample size selected for the Consultant A

[tex]n_{2}=5[/tex] sample size selected for the Consultant B

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

t 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 for the difference is equal to 0, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{1}- \mu_{2}=0[/tex]

Alternative hypothesis:[tex]\mu_{1} - \mu_{2}\neq 0[/tex]

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

[tex]t=\frac{6-7}{\sqrt{\frac{1^2}{5}+\frac{1.58^2}{5}}}=-1.195[/tex] (1)

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{1}+n_{2}-2=5+5-2=8[/tex]

Since is a two sided test the p value would be:

[tex]p_v =2*P(t_{(8)}<-1.195)=0.266[/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 FAIL to reject the null hypothesis. So the best option would be:

c. should not be rejected

Answer:

y=n*pi, where n is an integer.

Step-by-step explanation:

We have the equation:

[tex]y'(t)=3sin(y)[/tex]

Which is autonomous because the independent variable does not appear explicitly.

Now, we are asked about all the equilibrium solutions for such autonomous equation. Let's remember that the solution equilibrium y=y0 when F(y0)=0.

Then, matching the given equation to zero, we have:

[tex]y'(t)=3sin(y)=0[/tex]

Which is fulfilled when the sine function has a value of zero. The sine function is worth zero in y=n*pi where n is an integer.

Answer:

95% Confidence interval for the mean

[tex]142.8 \leq\mu\leq158.4[/tex]

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

[tex]cf=\sqrt{\frac{N-n}{N-1} }[/tex]

The standard deviation can be estimated as

[tex]\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963[/tex]

The 95% confidence interval has a z value of 1.96, so it becomes:

[tex]M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4[/tex]

Answer:

P-value = 0.1515

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 2.2

Sample mean, [tex]\bar{x}[/tex] = 2

Sample size, n = 52

Alpha, α = 0.05

Population standard deviation, σ = 1.4

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu =2.2\\H_A: \mu < 2.2[/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{2 - 2.2}{\frac{1.4}{\sqrt{52}} } = -1.03[/tex]

Now, we calculate the p-value from the normal standard table.

P-value = 0.1515

Answer:

The cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].

Step-by-step explanation:

It is given that the marginal cost of manufacturing an item when x thousand items are produced is

[tex]\frac{dC}{dx}=4x^3-6x+5[/tex]

We need to find the cost function.

Multiply both sides by dx.

[tex]dC=(4x^3-6x+5)dx[/tex]

Integrate both sides to find the cost function.

[tex]\int dC=\int (4x^3-6x+5)dx[/tex]

[tex]\int dC=4\int x^3dx-6\int xdx+5\int 1 dx[/tex]

[tex]C(x)=4(\frac{x^4}{4})-6(\frac{x^2}{2})+5x+C[/tex]

where, C(x) is const function and C is a constant.

[tex]C(x)=x^4-3x^2+5x+C[/tex]

It is given that C(0)=550. Substitute x=0 in the above function.

[tex]C(0)=(0)^4-3(0)^2+5(0)+C[/tex]

[tex]C(0)=C[/tex]

[tex]550=C[/tex]

Therefore, the cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].

Step-by-step explanation:

For an alternating series ∑a(n), the error bound of the sum of the first k terms is:

E < | a(k+1) |

Here, k = 10:

E < | a(11) |

E < | (-1)¹¹⁺¹ × 1 / (4(11) + 3) |

E < | 1/47 |

E < 1/47

Answer:

[tex]C(x) = \$8 + \$0.05*x[/tex]

Step-by-step explanation:

Let 'x' be the usage in kWh. When usage is at or under 800 kWh, the cost function C(x) is given by the base charge of $8 added to the rate of $0.05/kWh multiplied by the consumption 'x', in kWh.

Therefore, for 0 ≤ x ≤ 800, the cost function is:

[tex]C(x) = \$8 + \$0.05*x[/tex]

Answer:

She should add 30 ounces of the 12% vinegar she should add to 40 ounces of the 5% vinegar to obtain the needed vinegar with an 8% acidity level.

Step-by-step explanation:

Let X be ounces of the 12% vinegar she should add to 40 ounces of the 5% vinegar to obtain the needed vinegar with an 8% acidity level.

We have the equation

[tex]\frac{(X*0.12)+(40*0.05}{X+40} =0.08[/tex] when 8% acidity level is reached.

Solving the equation we have

X*0.12+2=0.08(X+40)=0.08X+3.2

That is 0.12X-0.08X=3.2-2=1.2

Finally 0.04X=1.2 and

X=30

Answer:

[tex]x=\frac{13}{5}, y= \frac{9}{5}[/tex]

Step-by-step explanation:

First Equation:

3x = y + 6

Or,

y = 3x - 6

2nd Equation:

x = 2y - 1

Lets substitute the expression for "y" from 1st equation in 2nd and solve for x:

x = 2y - 1

x = 2(3x - 6) - 1

x = 6x - 12 - 1

x = 6x - 13

5x = 13

x = 13/5

Now we can substitute this value of x into bold 1st equation and get y:

y = 3x - 6

y = 3(13/5) - 6

y = 9/5

This is the solution

Answer: 505

Step-by-step explanation:

The formula to find the sample size n , if the prior estimate of the population proportion (p) is known:

[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex] , where E=  margin of error and z = Critical z-value.

Let p be the population proportion of crashes.

Prior sample size = 250

No. of people experience computer crashes = 75

Prior proportion  of crashes [tex]p=\dfrac{75}{250}=0.3[/tex]

E= 0.04

From z-table , the z-value corresponding to 95% confidence interval = z=1.96

Required sample size will be :

[tex]n=0.3(1-0.3)(\dfrac{1.96}{0.04})^2[/tex] (Substitute all the values in the above formula)

[tex]n= (0.21)(49)^2= 0.21\times2401[/tex]

[tex]n= 504.21\approx505[/tex]   (Rounded to the next integer.)

∴ Required sample size = 505

Answer: above

Step-by-step explanation:

ok so angle is 30 and we have adjacent and we have to figure out opposite so adjacent and opposite, we need tan so

let x be height of triangle

tan 30= x/81

x= (tan 30) *81

x= about 48ft

48 + 4 is 52 which is more than 50

Answer:

[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 40,000 miles

Management believes that due to a new production process, the life expectancy of their tires has increased.

We design the null and the alternate hypothesis in the following manner:

[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]

The null hypothesis states before the new production process life expectancy of tires is equal to or less than 40,000 miles, while the alternate hypothesis states that after the new production process life expectancy of tires is greater than 40,000 miles.

Option D)

[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]

Final answer:

The correct hypotheses for checking if the average life expectancy of tires has increased are : μ ≤ 40,000 and : μ > 40,000, where μ represents the mean mileage for the population of tires. Option d) is the correct answer.

Explanation:

The correct set of hypotheses for testing whether the average life expectancy of tires produced by Whitney Tire Company has increased due to a new production process is:

μ ≤ 40,000 (null hypothesis)

μ > 40,000 (alternative hypothesis)

This hypothesis testing is set up to check if there is evidence to support the claim that the average life expectancy of the tires is greater than 40,000 miles. Thus, option d. : μ ≤ 40,000 : μ > 40,000 is correct. Here, μ stands for the mean mileage for the population of tires. The null hypothesis always contains an equality claim (≤, =, ≥), and the alternative hypothesis contains the opposite inequality, representing the claim we are trying to find evidence for.

Answer:

B.  the matched pairs t test.

Step-by-step explanation:

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=first test value , y = second test value

x: 920, 830, 960, 910

y: 1010, 800, 1000, 980

The system of hypothesis for this case are:

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

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

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

d: 90, -30, 40, 70

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{170}{4}=42.5[/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} =52.520[/tex]

The 4 step is calculate the statistic given by :

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{42.5 -0}{\frac{52.520}{\sqrt{4}}}=1.618[/tex]

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

[tex]df=n-1=4-1=3[/tex]

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

[tex]p_v =P(t_{(3)}>1.618) =0.1024[/tex]

So the p value is higher than any significance level assumed (0.05), so then we can conclude that we reject the null hypothesis. So we can conclude that the difference between the after and the initial score it's not significantly higher at 5% of significance.

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".  

Description in words of the parameter p

[tex]p[/tex] represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

[tex]\hat p=\frac{63}{200}=0.315[/tex] represent the estimated proportion of people that have experienced bullying

[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Numerical estimate for p

In order to estimate a proportion we use this formula:

[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.

[tex]\hat p=\frac{63}{200}=0.315[/tex] represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]  

For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

And replacing into the confidence interval formula we got:  

[tex]0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261[/tex]  

[tex]0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369[/tex]  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369

The lower limit of the confidence interval is 0.270, and the upper limit is 0.360.

To find the 90% z-confidence interval for the true proportion of students who have experienced bullying, we can use the formula:

CI = p ± z * sqrt((p(1-p))/n)

Where:

p is the sample proportion (63/200 = 0.315)

z is the z-value for a 90% confidence interval (z = 1.645)

n is the sample size (200)

Plugging in these values, we get:

CI = 0.315 ± 1.645 * sqrt((0.315(1-0.315))/200)

Simplifying the equation gives us:

CI = 0.315 ± 0.045

So, the lower limit of the confidence interval is 0.315 - 0.045 = 0.270, and the upper limit is 0.315 + 0.045 = 0.360.

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