34​% of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is​ (a) exactly​ two, (b) more than​ two, and​ (c) between two and​ five, inclusive.

Answer:

a) [tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]

b) [tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]

So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014

Step-by-step explanation:

For this case we assume the following model:

[tex]A(t)= 23.1 e^{0.0152 t}[/tex]

Where t is the number of years after 2000/

Part a

For this case we want the population for 2000 and on this case the value of t=0 since we have 0 years after 2000. If we rpelace into the model we got:

[tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]

So then the initial population at year 2000 is 23.1 million of people.

Part b

For this case we want to find the time t whn the population is 28.3 million.

So we need to solve this equation:

[tex]28.3= 23.1 e^{0.0152(t)}[/tex]

We can divide both sides by 23.1 and we got:

[tex]\frac{28.3}{23.1}= e^{0.0152t}[/tex]

Now we can apply natural log on both sides and we got:

[tex]ln(\frac{28.3}{23.1})= 0.0152 t[/tex]

And then for t we got:

[tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]

So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014

Answer:

a) The 90% confidence interval would be given by (721.716;790.724)  

b) We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)  

c) We assume the following conditions:

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".  

Part a

[tex]\bar X=756.22[/tex] represent the sample mean  

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

[tex]s=100.89[/tex] represent the sample standard deviation  

n=25 represent the sample size  

90% confidence interval  

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)  

The degrees of freedom are given by:

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

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,24)".And we see that [tex]t_{\alpha/2}=1.71[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]756.22-1.71\frac{100.89}{\sqrt{25}}=721.716[/tex]  

[tex]756.22+1.71\frac{100.89}{\sqrt{25}}=790.724[/tex]  

So on this case the 90% confidence interval would be given by (721.716;790.724)  

Part b

We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)  

Part c

We assume the following conditions:

Final answer:

The 90% confidence interval for the true speed of light is between 299,723.02683 km/sec and 299,789.41317 km/sec. This interval suggests we can be 90% confident that the constant speed of light falls within this range, with the understanding that the true speed of light is approximately 299,792,458 m/s.

Explanation:

The sample mean is 756.22, and the standard deviation is 100.89 with 25 trials.

First, add 299,000 km/sec to the sample mean to revert to the actual speed of light. Adjusted mean = 756.22 + 299,000 = 299,756.22 km/sec.

Since the sample size (25) is greater than 30, we use the z-score for a 90% confidence interval, which is 1.645.

To find the margin of error (ME), use the formula ME = z * (σ/√n), where σ is the standard deviation and n is the sample size. ME = 1.645 * (100.89/√25)= 1.645 * 20.178 = 33.19317 km/sec.

The confidence interval is then mean ± ME. That gives us the interval: [299,756.22 - 33.19317, 299,756.22 + 33.19317] or [299,723.02683, 299,789.41317] km/sec.

Interpretation: We are 90% confident that the true speed of light lies within the interval of 299,723.02683 to 299,789.41317 km/sec.

The samples are independent and randomly selected.

The data reported is accurate and measured without systematic errors.

The data is normally distributed or the sample size is large enough for the Central Limit Theorem to apply.

These results align with the known fact that the speed of light is a constant at approximately 299,792,458 meters/second, and any deviations observed in the experiment are likely due to measurement error or experimental uncertainties.

Final answer:

The equivalence class of 1 consists of all integers plus 1, and the equivalence class of 1/2 consists of all numbers of the form 1/2 plus any integer.

Explanation:

The equivalence relation R is defined such that (x, y) is in R if and only if x - y is an integer. For any real number a, the equivalence class of a is the set of all real numbers b such that a - b is an integer.

Equivalence Class of 1

The equivalence class of 1 includes all real numbers that are an integer distance from 1. This means it contains all numbers of the form 1 + k, where k is any integer. Hence, it includes numbers like 0, 1, 2, 3, and so on, in addition to negative integers: -1, -2, -3, etc.

Equivalence Class of 1/2

Similarly, the equivalence class of 1/2 consists of all real numbers of the form 1/2 + k, where k is any integer. This set includes numbers like -1/2, 1/2, 3/2, 5/2, and so on.

Answer:

[tex](0,0.75) \:or\:(0,\frac{3}{4})[/tex]

Step-by-step explanation:

Hi there!

1) Firstly, connect the points to draw a triangle.

2) From each vertex either with a pair of square or with a software trace a perpendicular line to the opposite side.

3) The concurrent point, i.e. the intersection point of these three altitudes is the orthocenter. Orthocenter means the the right center.

In equilateral triangles the Orthocenter coincides with the Centroid.

4) Finally, the coordinates of the Orthocenter found is (0,0.75)

Answer:

n>9

Step-by-step explanation:

4(n-3)-6>18

4n-12-6>18

4n-18>18

4n>18+18

4n>36

n>36/4

n>9

Answer: $49,875; $4,830C.

Step-by-step explanation:

The average salary of Teachers in a medium-sized suburban school district is $47,500 per year.

The standard deviation is $4,600

After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. The bonus of $500 will not alter the mean and standard deviation because equal amount is added for each teacher.

5% increase in each teacher's salary would differ. Therefore, it will affect the mean and standard deviation by 5%. Therefore, the new mean would be

47500 + (5/100 × 47500) = $49875

The new standard deviation would be

4600 + (5/100 × 4600) = $4830

Answer:

$50,375; $4,830 (I just answered it on one of my quizzes)

Step-by-step explanation:

Answer:

The probability of drawing a heart or diamond is 1/2 or 0.5

The probability that the card is not a spade is 3/4 or 0.75

Step-by-step explanation:

Consider the provided information.

Part (a) Find the probability of drawing a heart or diamond.

There are 13 cards of heart and 13 cards of diamond.

We need to find the probability of drawing a heart or diamond.

[tex]P(\text{Heart or Diamond})=P(\text{Heart card Drawn})+P(\text{Diamond card Drawn})[/tex]

[tex]P(\text{Heart or Diamond})=\frac{13}{52}+\frac{13}{52}[/tex]

[tex]P(\text{Heart or Diamond})=\frac{26}{52}=\frac{1}{2}=0.5[/tex]

Hence, the probability of drawing a heart or diamond is 1/2 or 0.5

(b) Find the probability that the card is not a spade.

Out of 52 cards 13 are spade,

That means 52 - 13 = 39 cards are not a spade.

[tex]P(\text{Not spade})=\frac{39}{52}=\frac{3}{4}=0.75[/tex]

Hence, the probability that the card is not a spade is 3/4 or 0.75

Answer:

a) [tex]\mu_1 -\mu_2[/tex] parameter of interest.

Where [tex]\mu_1[/tex] represent the mean response for adults

[tex]\mu_2[/tex] represent the mean response for teenegers

b) The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]

Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]

c) The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]

d) The 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]  

Step-by-step explanation:

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".  

Let group 1 be adults and group 2 be teenagers.

[tex]\bar X_1 =0.225[/tex] represent the sample mean 1

[tex]\bar X_2 =0.059[/tex] represent the sample mean 2

n1 represent the sample 1 size  

n2 represent the sample 2 size  

[tex]s_1 [/tex] sample standard deviation for sample 1

[tex]s_2 [/tex] sample standard deviation for sample 2

SE =0.091 represent the standard error for the estimate

(a) Give notation for the quantity that is being estimated.

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

(b) Give notation for the quantity that gives the best estimate.

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]

Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]

(c) Give the value for the quantity that gives the best estimate.

The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]

(d) Give a confidence interval for the quantity being estimated. Assuming 95% of confidence

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_1}+\frac{s^2_2}{n_2}}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]

We can assume that since we know the standard error the deviations are known and we can use the z distribution instead of the t distribution for the confidence interval.

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}=1.96[/tex]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=0.091[/tex]

Given by the problem

Now we have everything in order to replace into formula (1):  

[tex]0.166-1.96(0.091)=-0.012[/tex]  

[tex]0.166+1.96(0.091)=0.344[/tex]  

So on this case the 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]  

Final answer:

The quantity being estimated in the study is the difference in mean response to unlearn fear associations between adults and teenagers, denoted by Δμ = μ1 - μ2, where μ1 and μ2 represent the mean responses for adults and teenagers, respectively. This study contributes to understanding how fear associations are formed and unlearned, with implications on evolutionary predisposition towards certain fears.

Explanation:

The quantity being estimated in the study between adolescents and adults regarding their ability to unlearn fear associations tied to a specific sound is captured by the notation Δμ = μ1 - μ2. Here, μ1 represents the mean response for adults, and μ2 represents the mean response for teenagers. In this context, a higher physiological measure indicates less fear, with adults showing a mean response of 0.225 and teenagers showing a mean response of 0.059. The standard error of the estimate provided is 0.091, which helps in understanding the variability or precision of our estimated difference between the two groups' mean responses.

This study hints at the broader theory of preparedness, suggesting that humans are evolutionarily predisposed to easily associate certain stimuli with fear. Notably, the differentiation in fear response unlearning between age groups aligns with observations in social and developmental psychology about the specificity of fear acquisition and the challenges in modify these responses once established, especially during the teenage years.

Answer:

d) [-16.03,-3.97]

[tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].

Step-by-step explanation:

Notation and 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]n_A=18[/tex] represent the sample of A

[tex]n_B =13[/tex] represent the sample of B

[tex]\bar x_A =39[/tex] represent the mean sample  for A

[tex]\bar x_B =49[/tex] represent the mean sample for B  

[tex]s_A =8[/tex] represent the sample deviation for A

[tex]s_B =4[/tex] represent the sample deviation for B

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

Confidence =99% or 0.99

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

[tex](\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex] (1)  

The point of estimate for [tex]\mu_A -\mu_B[/tex] is just given by:  

[tex]\bar X_A -\bar X_B =39-49=-10[/tex]  

The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2=18+13-2=29[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that [tex]t_{\alpha/2}=2.756[/tex]  

The standard error is given by the following formula:  

[tex]SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex]  

And replacing we have:  

[tex]SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188[/tex]  

Confidence interval  

Now we have everything in order to replace into formula (1):  

[tex]-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03[/tex]  

[tex]-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97[/tex]  

So on this case the 99% confidence interval for the differences of means would be given by [tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].

d) [-16.03,-3.97]

The 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], therefore the correct answer is F. None of the above.

This question is about computing a confidence interval for the difference of two sample means. The formula for the 99% confidence interval for the difference between two means is:

(X1 - X2) ± Z * sqrt [s1^2/n1 + s2^2/n2]

Where X1 and X2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the Z-score for the desired confidence level. For a 99% confidence level, the Z-score is approximately 2.576. We plug the given values into the equation to calculate:

(39 - 49) ± 2.576 * sqrt [(8^2 / 18) + (4^2 / 13)] => -10 ± 2.576 * sqrt [3.56 + 1.23] => -10 ± 2.576 * sqrt [4.79] => -10 ± 2.576 * 2.19 => -10 ± 5.64

This means the 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], which is not among the given options, so the correct answer is F. None of the above.

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Answer:

[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]      

[tex]p_v =P(t_{15}>2)=0.0320[/tex]  

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.  

Step-by-step explanation:

1) Data given and notation      

[tex]\bar X=25[/tex] represent the sample mean      

[tex]s=2[/tex] represent the standard deviation for the sample      

[tex]n=16[/tex] sample size      

[tex]\mu_o =24[/tex] represent the value that we want to test    

[tex]\alpha[/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)  

Confidence =0.95 or 95%

[tex]\alpha=0.05[/tex]

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:      

Null hypothesis:[tex]\mu \leq 24[/tex]      

Alternative hypothesis:[tex]\mu > 24[/tex]      

We don't know the population deviation, so for this case 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{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]      

Calculate the P-value      

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

[tex]df=n-1=16-1=15[/tex]  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{15}>2)=0.0320[/tex]  

Conclusion      

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.      

Answer:

Standard error of the estimate of the difference between the two proportions=0.0259

Step-by-step explanation:

Given that the following data collected in two recent surveys of whether voters in cities A and B favor a ballot proposition in the next election.

City                         A                  B            Total

Sample size          615            585             1200

Favour X               463           403               866

Proportion p         0.7528     0.6889        0.7217

Std error for difference

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

p =0.7217

1-p = 0.2783

by substituting p and n1 = 615 and n2 = 585 we get

Std error = 0.0259

Standard error of the estimate of the difference between the two proportions=0.0259

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

The correct statement is that the middle 50% of personal injury claim costs fall between $989.72 and $1,088.18, which represents the interquartile range. This range and the standard deviation are key in evaluating the distribution of claim costs.

The statement that the middle 50% of the costs are between $989.72 and $1,088.18 is correct in reference to the provided summary statistics of personal injury claims. This range is defined by the first and third quartiles, also known as the interquartile range (IQR). The IQR is a measure of variability and represents the span between the 25th percentile (first quartile) and the 75th percentile (third quartile), which indeed encompasses the middle 50% of data in a given sample.

In the context of personal injury claims costs at your firm, this means that half of the claim costs fall within that range, with fewer costs being less than $989.72 (the lower 25%) and fewer costs being more than $1,088.18 (the upper 25%). This can be useful information for assessing claims costs and preparing for future claims expenses. The provided standard deviation of $89.50 indicates the average amount that claim costs vary from the mean ($1,040.47).

Answer:

8x + 6y >/= 48 ......1

x + 2y >/= 12 .......2

The cost function is given as;

C = 0.05x + 0.03y .........3

The minimum cost is $0.24 at (0,8)

That is 0 ounces of oats and 8 ounces of rice.

Step-by-step explanation:

let x represent the number of ounces of oats

And y represent the number of ounces of rice

For Vitamin A

Minimum requirements = 48mg

x ounces of oats contribute 8mg × x

y ounces of Rice contribute 6mg × y

Therefore, we have;

8x + 6y >/= 48 ......1

For Vitamin B

Minimum requirements = 12mg

x ounces of oats contribute 1mg × x

y ounces of Rice contribute 2mg × y

Therefore, we have;

x + 2y >/= 12 .......2

The cost function is given as;

C = 0.05x + 0.03y .........3

Attached is the graphical representation.

The feasible points are (x,y) = (0,8),(2.4,4.8),(12,0)

The minimum cost is determined by substituting each point into the cost function

For (0,8)

C= 0.05(0) + 0.03(8)

C = $0.24

For (12,0)

C= 0.60

For (2.4,4.8)

C= $0.264

The minimum cost is $0.24 at (0,8)

The problem can be modelled with a system of linear inequalities to represent the constraints of the cereal company. You graph these constraints and find the feasible region. After graphing the objective function, move this line towards the origin until it just leaves the feasible region. This point gives the optimal solution.

In this problem, we are dealing with linear equations and inequalities. The goal of the Munchies Cereal Company is to determine the amount of oats and rice, measured in ounces, to include in its cereal mix so as to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while at the same time minimizing the cost.

Let's denote the amount of oats as 'x' and the amount of rice as 'y'. The nutrition constraints can be formulated as:

And since quantities cannot be negative, we also have the constraints: x >= 0 and y >= 0. The objective is to minimize the cost, which can be expressed as C = 0.05x + 0.03y.

To solve this problem graphically, you would plot the constraint lines and see the feasible region (the area that satisfies all constraints). The cost line (C = 0.05x + 0.03y) is then drawn and moved towards the origin until the last point of the feasible region is touched. That point gives the optimal solution.

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Answer:

[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]

Step-by-step explanation:

Data given

[tex]\bar X=20[/tex] represent the sample mean for the sample  

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

[tex]s^2=0.02[/tex] represent the sample variance

[tex]s=0.141[/tex] represent the sample deviation

n=25 represent the sample size  

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 0.01, so the system of hypothesis are:

H0: [tex]\sigma \leq 0.1[/tex]

H1: [tex]\sigma >0.1[/tex]

In order to check the hypothesis we need to calculate the statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1=25-1=24 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 24 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,24)". And our critical value would be [tex]\chi^2 =36.415[/tex]

Since our calculated value is higher than the critical value we reject the null hypothesis at 5% of significance.

The variance hypothesis test for a cancer treatment drug with a sample mean of 20 mg and sample variance of 0.02 mg results in a chi-square test statistic of 48. This test statistic will be used to determine if the drug's variance exceeds the acceptable limit.

The question at hand is concerning a hypothesis test of the variance in dosage of a cancer treatment drug. The null hypothesis (H0) claims that the standard deviation of the drug content is not more than 0.1 mg, which corresponds to a variance of 0.01 mg² since variance = standard deviation². The alternative hypothesis (Ha) is that the variance is greater than 0.01 mg². Given the sample variance as 0.02 mg² and a sample size of 25, the test statistic for the chi-square test can be calculated using the formula:

Test statistic (chi-square) = (n - 1)*sample variance / hypothesized variance

Test statistic = (25 - 1) * 0.02 / 0.01 = 24 * 2 = 48

The calculated test statistic is 48. Since the sample variance is greater than the hypothesized variance, we have a test statistic that would fall in the rejection region based on the selected significance level in a Chi-square distribution, suggesting that the drug dosage may indeed have greater variability than the company's standard.

Answer:

The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.

Step-by-step explanation:

The volume of a cone is given by the following formula:

[tex]V = \frac{\pi r^{2}h}{3}[/tex]

We have that the diameter and the height are equal, so [tex]r = \frac{h}{2}[/tex]

So

[tex]V = \frac{\pi h^{3}}{12}[/tex]

Let's derivate this equation, using implicit derivatives.

[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]

In this problem, we have to:

Find [tex]\frac{dh}{dt}[/tex], when [tex]\frac{dV}{dt} = 20, h = 14[/tex]. So

[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]

[tex]20 = \frac{196\pi}{4}\frac{dh}{dt}[/tex]

[tex]\frac{dh}{dt} = \frac{20}{49\pi}[/tex]

The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.

This involves relationship between rates using Calculus.

dh/dt = 0.13 ft/min

Volumetric rate; dv/dt = 20 ft³/min

height of pile; h = 14 ft

We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.

Thus; diameter; d = 14 ft

radius; r = h/2 = d/2 = 14/2

radius; r= 7 ft

Plugging in the relevant values, we have;

20 = ¹/₄π × 14² × (dh/dt)

dh/dt = (20 × 4)/(π × 14²)

dh/dt = 0.13 ft/min

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We can use the inverse function derivative theorem:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=a} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(a)}}.[/tex]

In this case, we want to evaluate [tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1}[/tex], so:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}}.[/tex]

The derivative is:

[tex]\dfrac{\textrm{d}f}{\textrm{d}x} = \dfrac{\textrm{d}}{\textrm{d}x}\left[\ln(8x + \textrm{e})\right] = \dfrac{1}{8x+\textrm{e}}\dfrac{\textrm{d}}{\textrm{d}x}\left(8x + \textrm{e}\right) = \dfrac{8}{8x+\textrm{e}}.[/tex]

The ordinate of the point is [tex]f^{-1}(1) = 0[/tex], so we evaluate:

[tex]\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0} = \dfrac{8}{8 \times 0+\textrm{e}} = \dfrac{8}{\textrm{e}}.[/tex]

Finally:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0}} = \dfrac{1}{\dfrac{8}{\textrm{e}}} = \dfrac{\textrm{e}}{8}.[/tex]

We can check the answer by finding the inverse:

[tex]y = \ln(8x + \textrm{e}) \implies \textrm{e}^y = 8x + \textrm{e} \iff \textrm{e}^y - \textrm{e} = 8x \iff x = \dfrac{\textrm{e}^y-\textrm{e}}{8},[/tex]

so that

[tex]f^{-1}(x) = \dfrac{\textrm{e}^x-\textrm{e}}{8}.[/tex]

Therefore:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x} = \dfrac{\textrm{e}^x}{8}.[/tex]

Which finally gives the same answer as before:

[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}^1}{8} = \dfrac{\textrm{e}}{8}.[/tex]

Answer: [tex]\boxed{\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}}{8}}.[/tex]

Answer:

x = 4 + (log 6 / log 3)

x ≈ 5.631

Step-by-step explanation:

3^(x − 4) = 6

Take log base 3 of both sides.

log₃ 3^(x − 4) = log₃ 6

x − 4 = log₃ 6

Use change of base formula.

x − 4 = log 6 / log 3

Solve for x.

x = 4 + (log 6 / log 3)

x ≈ 5.631

Answer:

5.631

Step-by-step explanation:

Using the change of base formula log base b of y equals log y over log b

Log y (base b) = log y /log b

3^(x − 4) = 6

Taking the log of both sides

log 3^(x − 4) = log 6

using the logarithm law that states that

log a ^ x = x log a

x - 4 log 3 = log 6

x - 4 = log 6 / log 3

x - 4 = 1.630929754

x = 5.630929754

≈ 5.631

In a game of choice and payoff, Nick's dominant strategy is to choose 'Right', while Rosa lacks a dominant strategy. The Nash equilibrium is when Nick chooses 'Right' and Rosa chooses either 'Left' or 'Right' because changing their decisions would not lead to higher payoff.

The subject of this question is about a concept from game theory known as dominant strategies and the Nash equilibrium. Nick and Rosa are playing a game where they each simultaneously choose an action (Left or Right) and receive a payoff that depends on both their choices. To find the dominant strategy for each player, we need to identify what action that player would take, regardless of the other player's choice.

For Nick, the dominant strategy is to choose Right because his payoff (5 when Rosa picks Left, 6 when Rosa picks Right) is higher than when he picks Left (8 when Rosa picks Left, 4 when Rosa picks Right). For Rosa, she has no dominant strategy because her payoff is the same (4) whether she chooses Left or Right if Nick is choosing Left, and the same holds if Nick is choosing Right.

The Nash equilibrium is a situation where neither player can benefit by changing their strategy, assuming the other player stays the same. Here, the Nash equilibrium occurs when Nick chooses Right and Rosa chooses Left or Right, because neither player can gain a higher payoff by unilaterally changing their strategy.

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Answer:

False. The products from cross multiplication are different.

Step-by-step explanation:

To know if a pair of ratios form a proportion, cross multiply. If the products are equal, they are a proportion.

Write like this to see top (numerator) and bottom (denominator) clearly.

[tex]\frac{4}{5} =\frac{5}{6}[/tex]

Multiply each numerator with the other side's denominator:

4 X 6 = 24

5 X 5 = 25

Are they equal? No. 24 ≠ 25

Therefore it's not a proportion.

Answer: E , You will lose 5.6 cents

Step-by-step explanation:

Because with two dice, there are 36 possible outcomes, 21 are 7 or less, 14 are 8 through 11, and 1 is twelve.

Also when you have a total of 8,9,10 or 11, you gain $1 deducing the $1 you bet. The same with when you have 12 you gain $5.

Average $ to gain when total is 8,9,10 or 11 = P(8,9,10,11)

P(8,9,10,11) = ($2-$1)(14/36) = $14/36 gain

P(12) = ($6-$1)(1/36) = $5/36 gain

P(7 or less) = (0-$1)(21/36) = -$21/36 loss

P(loss or gain)= P(8,9,10,11) + P(12) + P(7 or less)

P(loss or gain) = $( 14/36 + 5/36 - 21/36) = -$2/36

P(loss or gain) = -$0.056 = -5.6 cents loss

Therefore, For every $1 bet you will lose 5.6 cents.

Answer: x = 120 degrees

Step-by-step explanation:

The diagram is that of a polygon with 5 sides. This means that it is a Pentagon. The sum of the interior angles in a polygon is expressed as

180(n -2)

Where n represents the number of sides that the polygon has.

Since the given polygon has 5 sides, then the sum of the interior angles would be

180(5 - 2) = 180 × 3 = 540 degrees.

Therefore,

x + x + x + 90 + 90 = 540

3x + 180 = 540

3x = 540 - 180 = 360

x = 360/3 = 120 degrees

Answer:

[tex]P(X\geq 2)=1-P(X\leq 1)=1-[0.054294+0.167762]=0.777944[/tex]

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]

Find the probability that at least 2 of the women sampled have been the victim of domestic abuse.

On this case we want to find this probability

[tex]P(X\geq 2) =1-P(X<2)=1-P(X\leq 1)= 1-[P(X=0)+P(X=1)][/tex]

And we can find the individual probabilities like this:

[tex]P(X=0)=(25C0)(0.11)^0 (1-0.11)^{25-0}=0.054294[/tex]  

[tex]P(X=1)=(25C1)(0.11)^1 (1-0.11)^{25-1}=0.167762[/tex]  

[tex]P(X\geq 2)=1-P(X\leq 1)=1-[0.054294+0.167762]=0.777944[/tex]

Using the binomial distribution, it is found that there is a 0.287825 = 28.7825% probability that at least 2 of the women sampled have been the victim of domestic abuse.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

In this problem:

The probability that at least 2 of the women sampled have been the victim of domestic abuse is given by:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which:

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

Hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{25,0}.(0.1111)^{0}.(0.8889)^{25} = 0.052641[/tex]

[tex]P(X = 1) = C_{25,1}.(0.1111)^{1}.(0.8889)^{24} = 0.164484[/tex]

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.052641 + 0.164484 = 0.217125[/tex]

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.217125 = 0.782875[/tex]

0.287825 = 28.7825% probability that at least 2 of the women sampled have been the victim of domestic abuse.

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Answer: Addie weighs 4 ounces

Missy weighs 14 ounces

Corky weighs 8 ounces

Step-by-step explanation:

Let a represent the weight of Addie.

Let m represent the weight of Missy.

Let c represent the weight of Corky.

Together Addie and Missy weigh 18 ounces. This means that

a + m = 18 - - - - - - - - - 1

Missy and Corky weigh 22 ounces. This means that

m + c = 22

m = 22 - c - - - - - - - - - - 2

Addie and Corky weigh 12 ounces. This means that

a + c = 12

a = 12 - c - - - - - - - - - - - 3

Substituting equation 2 and equation 3 into equation 1, it becomes

22 - c + 12 - c = 18

34 - 2c = 18

- 2c = 18 - 34 = - 16

c = - 16/ - 2 = 8

Substituting c = 8 into equation 2, it becomes

m = 22 - 8

m = 14

Substituting c = 8 into equation 3, it becomes

a = 12 - 8

a = 4

Answer:

[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]

And that would be the approximately cost for 2013.

Step-by-step explanation:

For this case we need to define some notation first.

A= population , t= represent the years after 2003, C= constant for the exponential model.

The starting point t=0 correspond to the year of 2003.

On this case we are assuming the following exponential model:

[tex]A(t) = A_o e^{Ct}[/tex]

The initial value on this case is for t=0 A(t=0)= 120 and if we replace we got this:

[tex]120=A_o e^{C(0)}=A_o e^0 = A_o[/tex]

And then the model is:

[tex]A(t) =120 e^{Ct}[/tex]

Now we need to determine the value for C. Since we know that inflation increase 4% per year we have that after one year we have 1.04 times the value of the original value, and we have this equation:

[tex]1.04 A_o= A_o e^{C(1)}= A_o e^C[/tex]

And we got this:

[tex]1.04= e^C [/tex]

Applying ln on both sides we got:

[tex]ln(1.04)= C=0.0392207[/tex]

So then our model is given by:

[tex]A(t) = 120 e^{ln(1.04)t}[/tex]

For 2013 we have that t=10 since 2013-2003 = 10 after 2003, if we replace t=10 we got this:

[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]

And that would be the approximately cost for 2013.

The cost of groceries in 2013, after applying an annual inflation rate of 4% for 10 years, will be approximately $177.63.

To calculate the annual rate of inflation and the cost in 2013, you can use the exponential growth function, where cost =[tex]initial_{cost} * (1 + rate)^{time[/tex]. In this case, the initial cost in 2003 (t=0) is $120 and the annual rate of inflation is 4%, or 0.04. To calculate the cost in 2013, which is 10 years after 2003, apply the exponential growth function:

Cost in 2013 = $120 * (1 + 0.04)¹⁰

This calculation yields:

Cost in 2013 = $120 * (1.04)¹⁰

Cost in 2013 = $120 * 1.48024

Cost in 2013 = $177.63

Therefore, in 2013, assuming the rate of inflation remains constant, the cart of groceries will cost approximately $177.63.

Answer:

Step-by-step explanation:

Answer:

a) The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:

[tex]\hat p=\frac{348}{400}=0.87[/tex]

b) [tex]np=400*0.87=348>10[/tex]

[tex]n(1-p)=400(1-0.87)=52>10[/tex]

So both conditions are satisifed and we can construct the confidence interval.

c) The 95% confidence interval would be given (0.837;0.903).

We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903

d) On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.

e) The 99% confidence interval would be given (0.827;0.913).

We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913

f) The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.

Step-by-step explanation:

Data given and notation  

n=400 represent the random sample taken    

X=348 represent the number of graduates that found jobs in the sample

[tex]\hat p=\frac{348}{400}=0.87[/tex] estimated proportion of graduates that found jobs

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

z would represent the statistic to calculate the confidence interval

p= population proportion of graduates that found jobs

(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?

The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:

[tex]\hat p=\frac{348}{400}=0.87[/tex]

(b) Check if the conditions for constructing a confidence interval based on these data are met.

[tex]np=400*0.87=348>10[/tex]

[tex]n(1-p)=400(1-0.87)=52>10[/tex]

So both conditions are satisifed and we can construct the confidence interval.

(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.96[/tex]

And replacing into the confidence interval formula we got:

[tex]0.87 - 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.837[/tex]

[tex]0.87 + 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.903[/tex]

And the 95% confidence interval would be given (0.837;0.903).

We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903

(d) What does "95% confidence" mean?

On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.

(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data

For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=2.58[/tex]

And replacing into the confidence interval formula we got:

[tex]0.87 - 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.827[/tex]

[tex]0.87 + 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.913[/tex]

And the 99% confidence interval would be given (0.827;0.913).

We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913

(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider?

The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.

Answer:

(a) The fraction of employees is 0.84.

(b)

[tex]\mu=850\\\\\sigma=450[/tex]

(c)

[tex]\mu=787.5\\\\\sigma=472.5[/tex]

(d) No. The left part of the distribution would be truncated too much.

Step-by-step explanation:

(a) If the weekly salaries are normally​ distributed, estimate the fraction of employees that make more than ​$300 per week.

We have to calculate the z-value and compute the probability

[tex]z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84[/tex]

(b) If every employee receives a​ year-end bonus that adds ​$100 to the paycheck in the final​ week, how does this change the normal model for that​ week?

The mean of the salaries grows $100.

[tex]\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850[/tex]

The standard deviation stays the same ($450)

[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2}  } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2}  }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2}  } =\sigma[/tex]

(c) If every employee receives a 5​% salary increase for the next​ year, how does the normal model​ change?

The increases means a salary X is multiplied by 1.05 (1.05X)

The mean of the salaries grows 5%, to $787.5.

[tex]\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5[/tex]

The standard deviation increases by a 5% ($472.5)

[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2}  } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2}  }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5[/tex]

(d) If the lowest salary is ​$300 and the median salary is ​$525​, does a normal model appear​ appropriate?

No. The left part of the distribution would be truncated too much.

Normal distribution has its mean, median and mode coincident on single point. The solutions to the given problems are specified as:

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z tables, the p value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

For the given case, if we take X = salary of employees weekly of the considered company, then:

[tex]X \sim N(750, 450)[/tex]

The  fraction of employees that make more than ​$300 per week is

P(X > 300).

Using the standard normal distribution, we can estimate this  fraction of employees that make more than ​$300 per week as:

[tex]P(X > 300 ) = 1 - P(X \leq 300) = 1 - P(Z = \dfrac{X - \mu}{\sigma} \leq \dfrac{300 - 750}{450} =-1)\\ \\P(X > 300) = 1 - P(Z \leq -1)\\[/tex]

Using the z-table, the p-value for Z = -1 is: 0.1587

Thus, [tex]P(X > 300) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413[/tex]

When we add $100 to each value of X, it doesn't change the structure of the graph of normal distribution.
When we increase salary by 5%, it means, new salary [tex]Y = X + 5\% \text{of X} = \dfrac{21X}{20}[/tex]

The random variable is scaled by 21/20 = 1.05, so the graph will stay on same origin but it will be stretched a bit in thickness and height.

For the 4th case(d), the median is specified to be $525, but as it was known to us that mean is $750, so mean and median aren't coinciding, and specially, the median is in left of mean, showing that the graph is leaning on right side(negatively skewed). (we deduce it when median < mean, as median shows that mid value is reached, but mean shows that probability is still not reached, so its being late, and reaches later, showing that there is tail in the left of the graph, so being negatively skewed).

Thus, The solutions to the given problems are specified as:

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Answer:

y = -4/6x - 4

Step-by-step explanation:

y = m(x - x₁) + y₁

You're given m=-4/6 and (0,-4) ←x₁=0, y₁=-4

so just plug it into the point-slope equation.

y = (-4/6)(x - (0)) + (-4)

y = (-4/6)(x) + (-4)

y = -4/6x - 4

Answer:y = -4x/6 - 4

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

The slope,m of the given line is -4/6

To determine the intercept, we would substitute m = -4/6, x = 0 and y = -4 into y = mx + c. It becomes

- 4 = -4/6 × 0 + c = 0 + c

c = - 4

The equation becomes

y = -4x/6 - 4

Answer:

The area of this revolted surface is 36π

Step-by-step explanation:

To obtain the area of a revolted surface, you have to define:

1) which is the axis on which the surface is revolted: this defines the limits on that axis or hight of the surface. In this case x∈[0;2]

2) which is the expression of the radius of the revolted surface and its dependence with the hight. In this case, the radius expression could be Y=4x+5

3) Define the angular variable: If this is a fully revolted surface, the angular variable will go from 0 to 2π

Now we can obtain the area with a double integral:

[tex]A=\int\limits^{2}_0 { \int\limits^{2\pi}_0 {r} \, d \varphi } \, dx =\int\limits^{2}_0 { \int\limits^{2\pi}_0 {4x+5} \, d \varphi } \, dx =\int\limits^{2}_0 { (2\pi)(4x+5)} \, dx=36\pi[/tex]

Answer:

A. 2 hours walking; 12 hours running

Step-by-step explanation:

The combination of hours walking and running has to respect both these inequalities:

[tex]3w + 6r \geq 36[/tex]

[tex]3w + 6r \leq 90[/tex]

A. 2 hours walking; 12 hours running

3w + 6r = 3*2 + 6*12 = 6+72 = 78.

Ok, it is larger than 35 and smaller than 91.

B. 4 hours walking; 3 hours running

3w + 6r = 3*4 + 6*3 = 12 + 18 = 30.

Invalid. Lesser than 36.

C. 9 hours walking; 12 hours running

3w + 6r = 3*9 + 6*12 = 27 + 72 = 99

Larger than 90. Invalid

D. 12 hours walking; 10 hours running

3w + 6r = 3*12 + 6*10 = 96

Larger than 90. Invalid

The combination of hours Keitaro can walk and run in a month to reach his goal is 2 hours walking; 12 hours running

3w + 6r ≥ 36. (1)

3w + 6r ≤ 90. (2)

substitute each option into the equation

A. 2 hours walking; 12 hours running

3w + 6r ≥ 36

3(2) + 6(12) ≥ 36

6 + 72 ≥ 36

78 ≥ 36

True

3w + 6r ≤ 90

3(2) + 6(12) ≤ 90

6 + 72 ≤ 90

78 ≤ 90

True

B. 4 hours walking; 3 hours running

3w + 6r ≤ 90

3(4) + 6(3) ≤ 90

12 + 18 ≤ 90

30 ≤ 90

True

B. 4 hours walking; 3 hours running

3w + 6r ≥ 36

3(4) + 6(3) ≥ 36

12 + 18 ≥ 36

30 ≥ 36

False

C. 9 hours walking; 12 hours running

3w + 6r ≥ 36

3(9) + 6(12) ≥ 36

27 + 72 ≥ 36

99 ≥ 36

True

3w + 6r ≤ 90

3(9) + 6(12) ≤ 90

27 + 72 ≤ 90

99 ≤ 90

False

D. 12 hours walking; 10 hours running

3w + 6r ≥ 36

3(12) + 6(10) ≥ 36

36 + 60 ≥ 36

96 ≥ 36

True

3w + 6r ≤ 90

3(12) + 6(10) ≤ 90

36 + 60 ≤ 90

96 ≤ 90

False.

Therefore, the combination of hours Keitaro can walk and run in a month to reach his goal is 2 hours walking; 12 hours running

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