Sean, a high school wrestler, has agreed to participate in a study of cardiovascular conditioning. He is left somewhat confused when, at the first research session, he is asked to complete a questionnaire about commonly purchased grocery items. Sean's confusion indicates a lack of ________ regarding the task. Question 2

Answer:

Step-by-step explanation:

a) We would apply the periodic interest rate formula which is expressed as

P = a/[{(1+r)^n]-1}/{r(1+r)^n}]

Where

P represents the monthly payments.

a represents the amount of the loan

r represents the annual rate.

n represents number of monthly payments. Therefore

a = $22000

r = 0.065/12 = 0.0054

n = 12 × 3 = 36

Therefore,

P = 22000/[{(1+0.0054)^36]-1}/{0.0054(1+0.0054)^36}]

22000/[{(1.0054)^36]-1}/{0.0054(1.0054)^36}]

P = 22000/{1.214 -1}/[0.0054(1.214)]

P = 22000/(0.214/0.0065556)

P = 22000/32.64

P = $674

b) The total amount that he would pay in 3 years is

674 × 36 = 24265

total interest paid over the life of the loan is

24265 - 22000 = $2265

Answer:

[tex] t=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]    

And the right critical value would be :  [tex]\Chi^2 =46.927[/tex]

And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 26 sample size

s= 0.18

[tex]\sigma_o =0.25[/tex] the value that we want to test

[tex]p_v [/tex] represent the p value for the test

t represent the statistic

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

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is equal to 0.25, so the system of hypothesis are:

H0: [tex]\sigma =0.25[/tex]

H1: [tex]\sigma \neq 0.25[/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 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=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]

And the right critical value would be :  [tex]\Chi^2 =46.927[/tex]

And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

At the 0.01 significance level, we fail to reject the null hypothesis, meaning there is not enough evidence to refute the claim that the standard deviation is 0.25.

To test the claim that the standard deviation of all precipitation amounts is equal to 0.25 using a 0.01 significance level, we use the chi-square test for standard deviation.

The null hypothesis (H0) states that the population standard deviation (σ) is 0.25. The alternative hypothesis (Ha) states that σ is not equal to 0.25.

The test statistic for a chi-square test is calculated using the formula:

Chi-square (χ²) = (n - 1)s² / σ₀²

Where:

Substituting the values:

χ² = (26 - 1)(0.18)² / (0.25)²

Calculating this:

χ² = 25 * 0.0324 / 0.0625 = 12.96

This χ² value is compared to the critical values from the chi-square distribution table with (n-1) or 25 degrees of freedom at a 0.01 significance level. The two-tailed critical values for 25 degrees of freedom at 0.01 significance level are approximately 10.85 and 46.93.

Since 10.85 < 12.96 < 46.93, we fail to reject the null hypothesis.

Hence, at the 0.01 significance level, there is not enough evidence to reject the claim that the standard deviation of all precipitation amounts is 0.25.

Answer:

C = 4T

Step-by-step explanation:

The question is not completed, the complete question and the solution is attached in the file below

Answer:

The future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually is $4,661.61

Step-by-step explanation:

The compound interest formula is given by:

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that

Semianually is twice a year, so [tex]n = 2[/tex].

Also, [tex]P = 1720, t = 14, r = 0.0725[/tex]

So

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

[tex]A = 1720*(1 + \frac{0.0725}{2})^{2*14}[/tex]

[tex]A = 4661.61[/tex]

The future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually is $4,661.61

The future value of $1,720 in 14 years, given a 7.25% interest rate compounded semiannually, can be calculated using the formula for compound interest: FV = PV * (1 + rate/n)^(nt). Plugging in the values, the future value comes out to be $3,490.91.

We can find the future value using the formula: Future Value = Present Value * (1 + rate/ n)^(n*t), where the Present Value is $1,720, the rate is 7.25% (or 0.0725 in decimal form), n is 2 (since it's compounded semiannually), and t, the time period, is 14 years.

So Future Value  =  $1,720 * (1 + 0.0725 / 2) ^ (2 * 14) .

Therefore, the future value of $1,720 in 14 years, with an interest rate of 7.25 percent compounded semiannually, would be $3,490.91

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Answer:

The probability table is shown below.

A Poisson distribution can be used to approximate the model of the number of hurricanes each season.

Step-by-step explanation:

(a)

The formula to compute the probability of an event E is:

[tex]P(E)=\frac{Favorable\ no.\ of\ frequencies}{Total\ NO.\ of\ frequencies}[/tex]

Use this formula to compute the probabilities of 0 - 8 hurricanes each season.

The table for the probabilities is shown below.

(b)

Compute the mean number of hurricanes per season as follows:

[tex]E(X)=\frac{\sum x f_{x}}{\sum f_{x}}=\frac{176}{68}= 2.5882\approx2.59[/tex]

If the variable X follows a Poisson distribution with parameter λ = 7.56 then the probability function is:

[tex]P(X=x)=\frac{e^{-2.59}(2.59)^{x}}{x!} ;\ x=0, 1, 2,...[/tex]

Compute the probability of X = 0 as follows:

[tex]P(X=0)=\frac{e^{-2.59}(2.59)^{0}}{0!} =\frac{0.075\times1}{1}=0.075[/tex]

Compute the probability of X = 1 as follows:

[tex]\neq P(X=1)=\frac{e^{-2.59}(2.59)^{1}}{1!} =\frac{0.075\times7.56}{1}=0.1943[/tex]

Compute the probabilities for the rest of the values of X in the similar way.

The probabilities are shown in the table.

On comparing the two probability tables, it can be seen that the Poisson distribution can be used to approximate the distribution of the number of hurricanes each season. This is because for every value of X the Poisson probability is approximately equal to the empirical probability.

To address this question, probabilities from the given data are calculated first, after which Poisson distribution is applied to compute the probabilities using the mean number of hurricanes per season. A comparison of both results offers an insight into the accuracy of the Poisson distribution in modeling this phenomenon.

To answer this question, we first have to calculate probabilities based on the empirical data and then compare them to probabilities computed under the assumption of a Poisson distribution.

First, we count total seasons from 1940 through 2007, which is 68. Using these frequencies, we can calculate the probability of having 0-8 hurricanes each season as follows: the number of seasons with a certain number of hurricanes divided by the total number of seasons.

For the Poisson distribution, we first need to calculate the average (mean) number of hurricanes per season, which is the sum of the product of the number of hurricanes and its frequency divided by the total number of seasons. After finding the mean, we can compute the probability of experiencing 0-8 hurricanes using the Poisson formula: e^(-mean) * (mean^n) / n!. After that, we can compare these probabilities with the ones derived from the empirical data.

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Answer:

a)Number of terms =15

Value of sum =

(b)Number of terms =6

Value of sum  

Step-by-step explanation:

A.

nth term,

a=5, r=

Simce the bases are the same, the powers are equal.

n-1=14

n=14+1=15

Number of terms =

15

Value of sum =

B.

nth term,

a=,

Simce the bases are the same, the powers are equal.

4+n-1=9

n=9+1-4=6

Number of terms =

6

Value of sum  

NOTE: For the sum, we desire an expression that gives the exact value, rather than entering an approximation

Answer:

The Answer is "G. P"

Step-by-step explanation:

They ask for "Which of the following vectors must be a PARTICULAR solution to the system?" so the answer simply G. which is P

Answer:

The value of given limit problem is 0.

Step-by-step explanation:

The given limit problem is

[tex]lim_{x\rightarrow \infty}\dfrac{6x^3+5x-7}{6x^4-4x^3-9}[/tex]

We need to find the value of given limit problem.

Divide the numerator and denominator by the leading term of the denominator, i.e., [tex]x^4[/tex]

[tex]lim_{x\rightarrow \infty}\dfrac{\frac{6x^3+5x-7}{x^4}}{\frac{6x^4-4x^3-9}{x^4}}[/tex]

[tex]lim_{x\rightarrow \infty}\dfrac{\frac{6}{x}+\frac{5}{x^3}-\frac{7}{x^4}}{6-\frac{4}{x}-\frac{9}{x^4}}[/tex]

Apply limit.

[tex]\dfrac{\frac{6}{ \infty}+\frac{5}{ \infty}-\frac{7}{ \infty}}{6-\frac{4}{ \infty}-\frac{9}{ \infty}}[/tex]

We know that [tex]\frac{1}{\infty}=0[/tex].

[tex]\dfrac{0+0-0}{6-0-0}[/tex]

[tex]\dfrac{0}{6}[/tex]

[tex]0[/tex]

Hence, the value of given limit is 0.

The limit of the expression (6x^3 + 5x - 7)/(6x^4 - 4x^3 - 9) as x approaches infinity is 0.

In mathematics, when we are asked to find the limit of an expression as x approaches infinity, we can use a technique known as the 'highest powers' method. This involves dividing all terms in the function by the highest power of x in the denominator. In this expression, the highest power of x in the denominator is x4. So, we'll divide all terms by x4.

The given expression is: (6x3+5x-7) / (6x4-4x3-9). Dividing all terms by x4, we get: (6/x+5/x3-7/x4) / (6-4/x-9/x4)

As x approaches infinity, all terms that have x in the denominator go to 0. So, our expression simplifies to (0/6) = 0. Hence, the limit of the given expression as x approaches infinity is 0.

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Final answer:

Mike's monthly car payment is approximately $489.99. The total cost of the car including interest over 60 months is $29,399.40, and the total interest Mike pays over the loan's lifetime is $4,399.40.

Explanation:

To determine Mike's monthly car payment, total cost of the car, and total interest paid over the life of the loan, we need to use the formula for the monthly payment on an installment loan, which can be found using the formula:

P = (Pv * r) / (1 - (1 + r)-n)

Where:

Lets calculate the monthly payment (P):

P = ($25,000 * 0.065/12) / (1 - (1+0.065/12)-60)
= $489.99 approximately

Mike's total cost of the car is the monthly payment multiplied by the number of payments:

$489.99 * 60 = $29,399.40

The total interest Mike pays is the total cost minus the loan amount:

$29,399.40 - $25,000 = $4,399.40

Thus, Mike's monthly payment is approximately $489.99, the total cost of his car will be $29,399.40, and the total interest paid over the life of the loan is $4,399.40.

Answer:

The average contents of all bottles of juice in the population

[tex]\mu = 473\text{ mL}[/tex]

Step-by-step explanation:

We are given the following in the question:

Population mean, μ =  473 mL

Sample mean, [tex]\bar{x}[/tex] = 472 mL

Sample size, n = 30

Sample standard deviation, s = 0.2

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 473\text{ mL}\\H_A: \mu < 473\text{ mL}[/tex]

Representation of [tex]\mu[/tex]

The average juice a bottle contain is the mean value of the juice.

[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 472mL.

The given parameters are:

[tex]\mathbf{n = 30}[/tex] --- the sample size

[tex]\mathbf{\sigma = 0.2}[/tex] --- the standard deviation

[tex]\mathbf{\bar x = 472}[/tex] --- the sample mean

[tex]\mathbf{\mu = 473}[/tex] --- the population mean

The above highlights means that:

The parameter [tex]\mathbf{\mu }[/tex] represents the population mean

This means that:

[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population.

From the question, the value is given as: 473

Hence, the true statement is:

[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 473mL.

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Answer:

0.6844 is the required probability.        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $1,250

Standard Deviation, σ = $125

We are given that the distribution of daily sales is a bell like shaped distribution that is a normal distribution.

Formula:

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

We have to find

P(sales less than $1,310)

[tex]P( x < 1310) = P( z < \displaystyle\frac{1310 - 1250}{125}) = P(z < 0.48)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 1310) =0.6844= 68.44\%[/tex]

0.6844 is the probability that sales on a given day at this store are less than $1,310.

The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]

It is given that, mean [tex]\mu=1250[/tex] and deviation [tex]\sigma=125[/tex]

The z- score is given as,

                   [tex]z-score=\frac{x-\mu}{\sigma} \\\\z=\frac{1310-1250}{125}=0.48 \\\\[/tex]

We have to find probability that the sales on a given day at this store are less than $1,310

                [tex]P(x < 1310)=P(z < 0.48)[/tex]

From z- value table.

           [tex]P(x < 1310)=68.44\%[/tex]

The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]

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Answer:

(a) The probability of catching 5 individuals in the pond is 0.1429.

(b) The probability of catching at least 2 individuals in the pond is 0.8838.

Step-by-step explanation:

Let X = number of water fleas caught by sweeping the water a single time.

The random variable X follows a Poisson distribution with parameter λ = 3.7.

The probability mass function of the Poisson distribution is:

[tex]P(X=x)=\frac{e^{-3.7}3.7^{x}}{x!}[/tex]

(a)

Compute the value of P (X = 5) as follows:

[tex]P(X=5)=\frac{e^{-3.7}3.7^{5}}{5!}=\frac{17.1443}{120} =0.142869\approx0.1429[/tex]

Thus, the probability of catching 5 individuals in the pond is 0.1429.

(b)

Compute the value of P (X ≥ 2) as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              [tex]=1-\frac{e^{-3.7}3.7^{0}}{0!}-\frac{e^{-3.7}3.7^{1}}{1!}\\=1-0.0247-0.0915\\=0.8838[/tex]

Thus, the probability of catching at least 2 individuals in the pond is 0.8838.

Answer:

The probability that the plane is oveloaded is P=0.9983.

The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.

Step-by-step explanation:

The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.

If the plane is full, we have 41 men in the plane. This is our sample size.

The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.

So the mean of the sample is 180.5 lb (equal to the population mean).

The standard deviation is:

[tex]\sigma=\frac{\sigma}{\sqrt{N}} =\frac{38.2}{\sqrt{41}}=\frac{38.2}{6.4} =5.97[/tex]

Then, we can calculate the z value for x=163 lb.

[tex]z=\frac{x-\mu}{\sigma}=\frac{163-180.5}{5.97}=\frac{-17.5}{5.97}= -2.93[/tex]

The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017

[tex]P(\bar X<163)=P(z<-2.93)=0.00169[/tex]

Then the probability that the plane is oveloaded is P=0.9983:

[tex]P(overloaded)=1-P(X<163)=1-0.0017=0.9983[/tex]

The pilot should take out the baggage or have less passengers in the plain to not overload.

Answer:

Marcus rents the car for 5 days

Step-by-step explanation:

For 5 days:

Plan A would equal $150 (30x5)

Plan B would equal $140 (125+15)

Plan b is cheaper only when Marcus rents the car for 5 days.

The question is not complete and the full question says;

Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

Answer:

A) Range = 30 minutes

B) Mean = 38

C) Mean Deviation = 7.2

D) This is well written in the explanation.

Step-by-step explanation:

A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.

Thus; Range = 54 - 24 = 30 minutes

B) In statistics,

Mean = Σx/n

Where n is the number of times occurring and Σx is the sum of all the times occurring

Thus,

Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380

n = 10

Thus, Mean(x') = 380/10 = 38

C) Mean deviation is given as;

M.D = [Σ(x-x')]/n

Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72

So, M.D = 72/10 = 7.2

D) The range of the times is 30 minutes.

The average time required to open one door is 38 minutes.

The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes

The average time to install an automatic garage door opener, based on the provided data, is approximately 38 minutes, though individual times can vary.

The subject of this question is Mathematics, specifically statistics. The question provides a series of data points representing the time, in minutes, that it took to install 10 automatic garage door openers. To find the average installation time, you would add up all of the times and then divide by the number of data points, which in this case is 10.

The data points are: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42. So, average installation time = (28+32+24+46+44+40+54+38+32+42) / 10 = 38 minutes.

This suggests that on average, it takes about 38 minutes to install an automatic garage door opener. But remember, this is an average. Individual installation times can vary greatly, as seen in the range of times provided.

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Answer:

(a) P(X>32.73) = 0.2912

(b) Out of 95 bottles, 28 would contain more than 32.73 oz of ale.

Step-by-step explanation:

(a) The amount of fill is normally distributed so we will calculate the z-score and then use it to find the probability using the normal distribution probability table.

Let the amount of fill be denoted by X. The z-score can be computed using the formula:

z = (X - μ)/σ

where μ = mean value of fill

          σ = standard deviation of value of fill = √Variance

P(X>32.73) = 1 - P(X<32.73)

                  = 1 - P((X-μ)/σ < (32.73 - 32.7)/√0.003)

                  = 1 - P(z<0.55)

Using the normal distribution table in Appendix B, we can see the probability at z=0.55 is 0.7088. So,

P(X>32.73) = 1 - 0.7088

P(X>32.73) = 0.2912

(b) We are buying 95 bottles and we need to calculate how many of them contain more than 32.73 oz. For that, we will multiply the total number of bottles by the probability of finding more than 32.73 oz which we have calculated in (a).

95 * 0.2912 = 27.664

Rounding off to a whole number we get 28 bottles.

Out of 95 bottles, 28 would contain more than 32.73 oz of ale.

Answer:

Incomplete question

Complete question: in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies "yes." You are interested in the number of freshmen you must ask. What is the probability that you will need to ask fewer than three freshmen?

Answer: P = 0.08656

Step-by-step explanation:

Using the concept of geometric distribution

P(X=x) = p(1-p)^1-x

Where x is the number of freshmen asked and p is the probability of success

Given probability of success as 71.3%

Therefore, P = 0.713

Probability that you'll need to ask less than 3 freshmen is given as

P = p(1-p)^1-x

x = 3 and p = 0.713

P = 0.713(1-0.713)^1-3

P = 0.713(0.287)^-2

P =0.713×12.140

P = 8.656%

P = 0.08656

Step-by-step explanation:

<y+60+58=180 (angle sum property)

<y+118=180

<y=180-118=62

<ACB+<ACD=180(degree of a line )

58+x=180

x=122

(btw u cant get a 100 points by answering one question)

Answer:

[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]

And we can use the complement rule and we got:

[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(1637.52,623.16)[/tex]  

Where [tex]\mu=1637.52[/tex] and [tex]\sigma=623.16[/tex]

Since the distribution of X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:

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

And we can use the z score formula given by:

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

And using this formula we got:

[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]

And we can use the complement rule and we got:

[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]

Answer: the probability is 0.49

Step-by-step explanation:

Since the account balances at the large bank are normally distributed.

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = account balances.

µ = mean account balance.

σ = standard deviation

From the information given,

µ = $1,637.52

σ = $623.16

We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as

P(x > 1650) = 1 - P(x ≤ 1650)

For x = 1650,

z = (1650 - 1637.52)/623.16 = 0.02

Looking at the normal distribution table, the probability corresponding to the z score is 0.51

P(x > 1650) = 1 - 0.51 = 0.49

Answer: the distance between the the police station and the fire station is 4 miles.

Step-by-step explanation:

Let x represent the distance between the library and the police station.

Let y represent the distance between the the police station and the fires station.

In the city, the distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station. This is expressed as

x = 2y - 3

The distance between the library and the police station is 5 miles. This means that

5 = 2y - 3

2y = 5 + 3 = 8

y = 8/2 = 4

The question given is incomplete, I googled and got the complete question as below:

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from over-fishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day. Use these data to address the following questions. Show your work.

a. What is the probability that your last pot will have the necessary 7 crabs?

b. What is the probability that your last pot will be empty?

Answer:

a. Probability = 0.0083

b. Probability = 0.0907

Step-by-step explanation:

This is Poisson distribution with parameter λ=2.4

a)

The probability that your last pot will have the necessary 7 crabs is calculated below:

P(X=7)=  {e-2.4*2.47/7!} = 0.0083

b)

The probability that your last pot will be empty is calculated as:

P(X=0)=  {e-2.4*2.40/0!} = 0.0907

The probability of getting 7 crabs in the last pot is approximately 0.0082, while the probability of the last pot being empty is roughly 0.0907.

To solve this problem, we need to use the Poisson distribution. The Poisson distribution is a probability distribution that can be used to predict the number of events occurring within a fixed interval of time or space. Given that the average (λ) number of crabs per pot is 2.4, we can proceed to solve for the probabilities.

a. Probability of having 7 crabs in the last pot:

b. Probability of the last pot being empty:

The complete question is

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from overfishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day

Use these data to address the following questions. Show your work.

a. What is the probability that your last pot will have the necessary 7 crabs?

b. What is the probability that your last pot will be empty?

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

The conditional distribution of income level among single men should add up to 100%, since when calculating a conditional distribution we treat the subgroup as the total population. Thus, income levels among single men are all the possible outcomes for this group and should collectively correspond to 100% of cases.

The conditional distribution in this scenario is the distribution of income level among single men, being key to understanding he relative frequency of various income levels within this specific group. Since we are looking at a subset of the total population (in this case, single men), the percentages should indeed add up to 100%. This is because when we calculate the conditional distribution, we treat this subgroup as if it were the total population. Therefore, the total should be 100%, reflecting the entire subgroup, not compared to the entire population. An example could be: If 10% of single men are in low income, 30% are in middle income, and 60% are in high income, these percentages represent the income level distribution among single men, and add up to 100%.

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Based on the properties of the Normal Distribution and Standard Deviations, we find that about 0.3% of the items will weigh either less than 87 grams or more than 93 grams.

To solve this question, we need to understand and apply concepts of the Normal Distribution and Standard Deviations. A normal distribution is a common type of statistical distribution representing various types of data, characterized by a bell-shaped curve symmetric about its mean.

In this case, the item weights are normally distributed with a mean (average) of 90 grams and a standard deviation of 1 gram. Standard deviation basically tells us how much the data varies around the mean value.

The question now asks us to find out what percentage of weights are either less than 87 grams or more than 93 grams. We know that within 1 standard deviation of the mean (between 89 and 91 grams in this case), we have about 68.2% of the data. Similarly, within 2 standard deviations of the mean (between 88 and 92 grams), we have about 95.4% of the data. And within 3 standard deviations (between 87 and 93 grams), we have about 99.7% of the data.

So, since 87 grams and 93 grams are 3 standard deviations away from the mean, we know that about 99.7% of the weights lie between these two values. This means that the other 0.3%, or 0.15% on either side, is either below 87 grams or above 93 grams. Hence, 0.3% of the items will either weigh less than 87 grams or more than 93 grams.

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Answer:

[tex]\frac{dy}{dx} = 2x ( 2cos2x) +sin2x[/tex]

[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]

Step-by-step explanation:

Given y = x sin2x .....(1)

Applying UV formula [tex]\frac{d(UV)}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}[/tex]

Differentiating with respective to 'x' we get

[tex]\frac{dy}{dx} = x ( 2cos2x)\frac{d(2x)}{dx} +sin2x (1)[/tex]

[tex]\frac{dy}{dx} = x ( 2cos2x)(2) +sin2x (1)[/tex]

Final answer:-

[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]

Answer:

a. A = {1, 2, 3, 4, 5, 6}

b. B = {1, 2, 3, 4, 5, 6}

c. C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

d. D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

Step-by-step explanation:

a. the maximum value to appear in the two rolls

Since only the maximum value is computed, the variable can assume any integer from 1 to 6:

A = {1, 2, 3, 4, 5, 6}

b. the minimum value to appear in the two rolls;

Since only the minimum value is computed, the variable can assume any integer from 1 to 6:

B = {1, 2, 3, 4, 5, 6}

c. the sum of the two rolls;

The minimum value would be from rolling two ones (sum is 2) and the maximum value would be from rolling two sixes (sum is 12). Every integer in-between is possible:

C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

d. the value of the first roll minus the value of the second roll

The minimum value would be from rolling a one and a six (result is -5) and the maximum value would be from rolling a six and a one (result is 5). Every integer in-between is possible:

D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

Final answer:

Explanation of possible values for maximum, minimum, sum, and difference in rolling two dice.

Explanation:

a. The possible values for the maximum value:

1, if the same number appears twice on both rolls

2 to 6, for individual numbers in different scenarios

b. The possible values for the minimum value:

1 to 5, as the minimum value will exclude the maximum value

c. The possible values for the sum:

2 to 12, representing all possible sums from rolling two dice

d. The possible values for the difference between two rolls:

-5 to 5, as the difference can range from -5 to 5

Answer:

Below is the R code for the bootstrapping in exponential distribution. The result is attached below.

####################################

rm(list=ls(all=TRUE))

set.seed(12345)

N=c(10,100,500)

Rate=0.2

B=1000

MN=SE=rep()

for(i in 1:length(N))

{

n=N[i]

X=rexp(n,rate=Rate)

EST=1/mean(X)

ESTh=rep()

for(j in 1:B)

{

Xh=rexp(n,rate=EST)

ESTh[j]=1/mean(Xh)

}

MN[i]=mean(ESTh)

SE[i]=sd(ESTh)

}

cbind(N,Rate,MN,SE)

Step-by-step explanation:

Answer:

16.35% probability their combined weight exceeds 46000 pounds.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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

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

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

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

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 1135, \sigma = 97, n = 40, s = \frac{97}{\sqrt{40}} = 15.34[/tex]

Find the probability their combined weight exceeds 46000 pounds.

This is 1 subtracted by the pvalue of Z when [tex]X = \frac{46400}{40} = 1150[/tex]. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{1150 - 1135}{15.34}[/tex]

[tex]Z = 0.98[/tex]

[tex]Z = 0.98[/tex] has a pvalue of 0.8365

1 - 0.8365 = 0.1635

16.35% probability their combined weight exceeds 46000 pounds.

Final answer:

The probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds is approximately 16.35%. This calculation uses the Z-score related to the sample mean weight exceeding 1,150 pounds.

Explanation:

To find the probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds, we first need to determine if the average weight per bull exceeds 1,150 pounds (since 46,000 ÷ 40 = 1,150). The mean weight of a Hereford bull is given as 1,135 pounds with a standard deviation of 97 pounds. Using the Central Limit Theorem, we can find the mean and standard deviation for the sample mean weight of 40 bulls.

The standard deviation of the sample mean (σ_x-bar) is σ ÷ √{n} = 97 pounds ÷ √{40}, which equals approximately 15.34 pounds. To find the Z-score, we use the formula Z = (X - μ) ÷ σ_x-bar, where X is 1,150 pounds, μ (the mean) is 1,135 pounds, and σ_x-bar is 15.34 pounds, yielding a Z-score of about 0.98.

Using the standard normal distribution table, a Z-score of 0.98 corresponds to a probability of approximately 0.8365. However, since we want the probability that the sample mean is greater than 1,150 pounds, we need to look at the upper tail, which is 1 - 0.8365 = 0.1635. Therefore, there's a 16.35% chance the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds.

Answer:

[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]  

[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]  

And the 95% confidence interval would be given (0.0181;0.181).  

We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/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".  

Solution to the problem

[tex]p_A[/tex] represent the real population proportion for telephone  

[tex]\hat p_A =0.28[/tex] represent the estimated proportion for telephone

[tex]n_A=200[/tex] is the sample size required for telephone

[tex]p_B[/tex] represent the real population proportion for mail

[tex]\hat p_B =0.18[/tex] represent the estimated proportion for mail

[tex]n_B=200[/tex] is the sample size required for mail

[tex]z[/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]  

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/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](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]  

[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]  

And the 95% confidence interval would be given (0.0181;0.181).  

We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/tex]

The 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail is between 1.14% and 18.86%.

To calculate a 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail, we use the formula for comparing two proportions.

We are given that 28% of the 200 contacted by telephone donated, which is a proportion of p1 = 0.28, and 18% of the 200 contacted by first class mail donated, which is a proportion of p2 = 0.18. Each group size is n1 = n2 = 200.

The formula for the standard error of the difference between two independent proportions is:

SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)

So, the standard error (SE) for our example is:

SE = sqrt(0.28(1-0.28)/200 + 0.18(1-0.18)/200) = sqrt(0.001248 + 0.000792) = sqrt(0.00204)

The Z-score for a 95% confidence interval is approximately 1.96. Therefore, the margin of error (MOE) is calculated as:

MOE = 1.96 * SE

We then compute the confidence interval as:

(p1 - p2) ± MOE = (0.28 - 0.18) ± 1.96 * sqrt(0.00204)

Calculate the MOE:

MOE = 1.96 * sqrt(0.00204) ≈ 1.96 * 0.0452 ≈ 0.0886

Then the 95% confidence interval is:

(0.10 ± 0.0886) which is (0.0114, 0.1886)

This means that we are 95% confident that the true difference in the proportion of people who would donate when contacted by telephone versus first class mail is between 1.14% and 18.86%.

Answer: LAST OPTION.

Step-by-step explanation:

For this exercise it is important to remember that, by definition, the angle formed by two secants intersecting outside of a circle, can be found with the following formula:

[tex]Angle\ formed\ by\ two\ secants=\frac{Difference\ of\ intercepted\ arcs}{2}[/tex]

According to the information provided in the exercise, you know that:

[tex]mAC= 93\°\\\\mBD= 39\°[/tex]

Therefore, you can find the difference of the intercepted arcs:

[tex]Difference\ of\ intercepted\ arcs=93\°-39\°\\\\Difference\ of\ intercepted\ arcs=54\°[/tex]

The final step is to substitute  the difference of the intercepted arcs calculated above, into the formula, in order to find the measure of the angle  BED.

You get that this is:

[tex]Angle\ formed\ by\ two\ secants=\frac{54\°}{2}\\\\Angle\ formed\ by\ two\ secants=27\°[/tex]

Answer:

The answer should be 27

Step-by-step explanation:

Answer:

a) 81.85% of employees put between $9, 500 and $11,000 into the 401 k per year

b) 0.13% of employee put more than $11, 500 into the 401 k per year

c) 97.72% of employees put less than $11,000 into the 401k per year.

d) 97.72% of employees put more than $9,000 into the 401k per year

e) 2.15% of employees put between than $11,000 and $11, 500 into the 401k per year

f) An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors

Step-by-step explanation:

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

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

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

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

In this problem, we have that:

[tex]\mu = 10000, \sigma = 500[/tex]

a. what proportion of employees put between $9, 500 and $11,000 into the 401 k per year

This is the pvalue of Z when X = 11000 subtracted by the pvalue of Z when X = 9500. So

X = 11000

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

[tex]Z = \frac{11000 - 10000}{500}[/tex]

[tex]Z = 2[/tex]

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

X = 9500

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

[tex]Z = \frac{9500 - 10000}{500}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

0.9772 - 0.1587 = 0.8185

81.85% of employees put between $9, 500 and $11,000 into the 401 k per year

b. What proportion of employee put more than $11, 500 into the 401 k per year?

This is 1 subtracted by the pvalue of Z when X = 11500. So

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

[tex]Z = \frac{11500 - 10000}{500}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% of employee put more than $11, 500 into the 401 k per year

c. What proportional of employees put less than $11,000 into the 401k per year?

This is the pvalue of Z when X = 11000. So

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

[tex]Z = \frac{11000 - 10000}{500}[/tex]

[tex]Z = 2[/tex]

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

97.72% of employees put less than $11,000 into the 401k per year.

d. What proportional of employees put more than $9,000 into the 401k per year?

This is 1 subtracted by the pvalue of Z when X = 9000. So

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

[tex]Z = \frac{9000 - 10000}{500}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228.

1 - 0.0228 = 0.9772

97.72% of employees put more than $9,000 into the 401k per year

e. What proportional of employees put between than $11,000 and $11, 500 into the 401k per year?

This is the pvalue of Z when X = 11500 subtracted by the pvalue of Z when X = 11000. So

X = 11500

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

[tex]Z = \frac{11500 - 10000}{500}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a pvalue of 0.9987

X = 11000

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

[tex]Z = \frac{11000 - 10000}{500}[/tex]

[tex]Z = 2[/tex]

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

0.9987 - 0.0972 = 0.0215

2.15% of employees put between than $11,000 and $11, 500 into the 401k per year

f. How much would an employees need to put into his or her 401 K to be in the upper 10% of investors?

This is the value of Z when X has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28.

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

[tex]1.28 = \frac{X - 10000}{500}[/tex]

[tex]X - 10000 = 500*1.28[/tex]

[tex]X = 10640[/tex]

An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors