The Customer Service Center in a large New York department store has determined that the amount of time spent with a customer about a complaint is normally distributed, with a mean of 9.3 minutes and a standard deviation of 2.6 minutes. What is the probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be as follows. (Round your answers to four decimal places.)

Answers

Answer 1

Answer:

a) 0.6062

b) 0.9505

c) 0.679

Step-by-step explanation:

The customer service center in a large new york department store has determined tha the amount of time spent with a customer about a complaint is normally distributed, with a mean of 9.3 minutes and a standard deviation of 2.5 minutes. What is the probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be

(a) less than 10 minutes?

(b) longer than 5 minutes?

(c) between 8 and 15 minutes?

a) The Z score (z) is given by the equation:

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

Where:

μ is the mean = 9.3 minutes,

σ is the standard deviation = 2.6 minutes and x is the raw score

[tex]z=\frac{x-\mu}{\sigma}=\frac{10-9.3}{2.6}=0.27[/tex]

From the z tables, P(X < 10) = P(z < 0.27) = 0.6062 = 60.62%

b) The Z score (z) is given by the equation:

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

[tex]z=\frac{x-\mu}{\sigma}=\frac{5-9.3}{2.6}=-1.65[/tex]

From the z tables, P(X > 5) = P(z > -1.65) = 1 - P(z < -1.65) = 1 - 0.0495 =  0.9505 = 95.05%

c) For 8 minutes

[tex]z=\frac{x-\mu}{\sigma}=\frac{8-9.3}{2.6}=-0.5[/tex]

For 15 minutes

[tex]z=\frac{x-\mu}{\sigma}=\frac{15-9.3}{2.6}=2.19[/tex]

From the z tables, P(8< X < 15) = P(-0.5 < z < 2.19) = P(z < 2.19) - P(z< -0.5) = 0.9875 - 0.3085 = 0.679 = 67.9%

Answer 2
Final answer:

The question pertains to finding a probability concerning the time taken to resolve a customer's complaint. The time follows a normal distribution with a mean of 9.3 minutes and a standard deviation of 2.6 minutes. We need to calculate the Z-score with the required time, mean and standard deviation, which can then be referenced on a standard normal distribution table to find the probability.

Explanation:

The question is about finding the probability of the time spent with a customer in a Customer Service Center of a department store in New York, given that the time that is spent follows a normal distribution with a mean of 9.3 minutes, and a standard deviation of 2.6 minutes.

To calculate this, we'll use the standard normal distribution, Z-score, which standardizes the distribution. The Z-score is a measure of how many standard deviations an element is from the mean. It can be calculated by using the formula: Z = (X - μ) / σ

where X is the time about which we want to find the probability, μ is the mean, and σ is the standard deviation.

Unfortunately, the exact time (X) you want to find the probability for was not provided in your question. However, assuming X to be a given time, t, you substitute t, 9.3 (the mean), and 2.6 (the standard deviation) into the formula to get a Z-score. Then, by referencing the Z-score on a standard normal distribution table, you can find the probability for a complaint taking at an amount of time, t, to be resolved.

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Related Questions

You spin the spinner shown below once. Each sector shown has an equal area.

What is p (not squirrel)

Answers

Answer: 0.8

Step-by-step explanation:

There are 5 possible equally sized sections meaning there is a 0.2 or 20% of landing on each section. We can subtract the probability of getting a squirrel from the whole thing (1 as probability always equals 1) to determine the probability of not squirrel.

P(squirrel) = 0.2

1 - P(squirrel) = P(not a squirrel)

1 - 0.2 = 0.8

Answer: the anserw is 0.8

Step-by-step explanation:

We survey a random sample of American River College students and ask if they drink coffee on a regular basis. The 90% confidence interval for the proportion of all American River College students who drink coffee on a regular basis is (0.262, 0.438). What will be true about the 95% confidence interval for these data? Group of answer choices The 95% confidence interval is narrower than the 90% confidence interval. The 95% confidence interval is wider than the 90% confidence interval. The two intervals will have the same width. It is impossible to say which interval will be wider.

Answers

Answer:

Correct option:

The 95% confidence interval is wider than the 90% confidence interval.

Step-by-step explanation:

The (1 - α)% confidence interval for the population proportion is:

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

The width of this interval is:

[tex]W=UL-LL\\=2\times z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

The width of the interval is directly proportional to the critical value.

The critical value of a distribution is based on the confidence level.

Higher the confidence level, higher will be critical value.

The z-critical value for 95% and 90% confidence levels are:

[tex]90\%: z_{\alpha /2}=z_{0.05}=1.645\\95\%: z_{\alpha /2}=z_{0.025}=1.96[/tex]

*Use a z-table.

The critical value of z for 95% confidence level is higher than that of 90% confidence level.

So the width of the 95% confidence interval will be more than the 90% confidence interval.

Thus, the correct option is:

"The 95% confidence interval is wider than the 90% confidence interval."

Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random sample of 50 students, she found that 32 students read at least 1 book last month. Assuming all conditions for Inference are met, which of the following defines a 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month? 32 +1.645,8908 32 + 1.9620 0.64 + 1.282,16.620.) © 264 265 CHAT 0.04 + 1.00, layanan

Answers

Answer:

0.64 ± 0.1117 or

[tex]0.64\pm 1.645*\sqrt{\frac{0.64*(0.36)}{50}}[/tex]

Step-by-step explanation:

Sample size (n) = 50

Z-score for a 90% confidence interval (z) = 1.645

Proportion of students that read at least one book (p):

[tex]p=\frac{32}{50}=0.64[/tex]

The confidence interval is given by:

[tex]p\pm z*\sqrt{\frac{p*(1-p)}{n} }[/tex]

Applying the given data:

[tex]0.64\pm 1.645*\sqrt{\frac{0.64*(1-0.64)}{50}}\\ 0.64\pm 0.1117[/tex]

The confidence interval is 0.64 ± 0.1117

The 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month is [tex]0.64\pm 0.11178[/tex].

Given information:

From a random sample of 50 (n) students, Alma found that 32 students read at least 1 book last month.

Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book.

It is required to find the 90 percentage confidence interval for the proportion of all students in her district who read at least 1 book last month.

Now, from tha table, the z-score of 90 percent confidence level is,

[tex]z=1.645[/tex]

The probability p for the proportion of students who read atleast one book is,

[tex]p=\dfrac{32}{50}\\p=0.64[/tex]

So, the 90 percent confidence interval will be calculated as,

[tex]p\pm z\sqrt{\dfrac{p(1-p)}{n}}=0.64\pm 1.645\times \sqrt{\dfrac{0.64(1-0.64)}{50}}\\=0.64\pm 0.11178[/tex]

Therefore, the 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month is [tex]0.64\pm 0.11178[/tex].

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please help me ASAP!!

Answers

Answer:

x  = 16

Step-by-step explanation:

The sum of measures of these three angles of any triangle is equal to 180 °

54 + 90 + (2x + 4) = 180

144 + 2x + 4 = 180

148 + 2x = 180

2x = 180 - 148

2x = 32

[tex]x= \frac{32}{2} \\\\x=16[/tex]

Focus on triangle KLM. Since JKLM is a rectangle, angle KLM is right. Since the sum of the interior angles of every triangle is 180, we deduce that

[tex]\hat{K}+\hat{L}+\hat{M}=180[/tex]

Now plug the values for each angle and you have

[tex]54+90+(2x+4)=180[/tex]

Move 54 and 90 to the right side to get

[tex]2x+4=36 \iff 2x=32 \iff x=16[/tex]

The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.

Answers

Answer:

(a) [tex]X\sim N(0.10,\ 0.02^{2})[/tex].

(b) The probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.

(c) The 60th percentile is 11%.

Step-by-step explanation:

The complete question is:

The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.  

A. X ~ N( , )  

B. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11%.

C.Find the 60th percentile for this distribution.

Solution:

The random variable X is defined as the THC content for a bag of marijuana that is sold on the street.

(a)

The random variable X is Normally distributed with mean, μ = 0.10 and standard deviation, σ = 0.02.

Thus, [tex]X\sim N(0.10,\ 0.02^{2})[/tex].

(b)

Compute the value of P (X > 0.11) as follows:

[tex]P(X>0.11)=P(\frac{X-\mu}{\sigma}>\frac{0.11-0.10}{0.02})\\=P(Z>0.50)\\=1-P(Z<0.50)\\=1-0.69146\\=0.30854\\\approx 0.31[/tex]

*Use a z-table.

Thus, the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.

(c)

The pth percentile is a data value such that at least p% of the data-set is less than or equal to this data value and at least (100 - p)% of the data-set are more than or equal to this data value.

Let x represent the 60th percentile value.

The, P (X < x) = 0.60.

⇒ P (Z < z) = 0.60

The value of z for this probability is,

z = 0.26.

Compute the value of x as follows:

[tex]z=\farc{x-\mu}{\sigma}\\0.26=\farc{x-0.10}{0.02}\\x=0.10+(0.26\times 0.02)\\x=0.1052\\x\approx 0.11[/tex]

Thus, the 60th percentile is 11%.

Using the common denominator what is the equivalent fraction of 1/8

Answers

Answer:

2/16

Step-by-step explanation:

1/8 x 2 = 2/16

Both 2 and 16 divide by 8, so one can know this is correct.

Also, both 8 and 16 are divisible by 8, so one can see this is the common denominator.

Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8) (a) the tangent plane (b) the normal line (x(t), y(t), z(t)) =

Answers

Let [tex]f(x,y,z)=x+y+z-8e^{xyz}[/tex]. The tangent plane to the surface at (0, 0, 8) is

[tex]\nabla f(0,0,8)\cdot(x,y,z-8)=0[/tex]

The gradient is

[tex]\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)[/tex]

so the tangent plane's equation is

[tex](1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8[/tex]

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by [tex]t[/tex], then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

[tex](1,1,1)t+(0,0,8)=(t,t,t+8)[/tex]

or [tex]x(t)=t[/tex], [tex]y(t)=t[/tex], and [tex]z(t)=t+8[/tex].

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)

Final answer:

This detailed answer explains how to find the equation of the tangent plane and the normal line to a given surface at a specified point. The parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]

Explanation:

To find the equations of the tangent plane and the normal line to the given surface x + y + z = 8exyz at the specified point 0, 0, 8) we first need to find the partial derivatives of x + y + z - 8exyz with respect to x, y, and z.

Given the surface equation:

[tex]\[ x + y + z = 8exyz \][/tex]

Let's denote the function as [tex]\(F(x, y, z) = x + y + z - 8exyz\).[/tex]

(a) To find the equation of the tangent plane, we use the gradient vector of F at the point 0, 0, 8 as the normal vector to the tangent plane.

The gradient vector of F is given by:

[tex]\[ \nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \][/tex]

Let's calculate the partial derivatives:

[tex]\[ \frac{\partial F}{\partial x} = 1 - 8eyz \][/tex]

[tex]\[ \frac{\partial F}{\partial y} = 1 - 8exz \][/tex]

[tex]\[ \frac{\partial F}{\partial z} = 1 - 8exy \][/tex]

Now, evaluate these partial derivatives at the point (0, 0, 8):

[tex]\[ \frac{\partial F}{\partial x} = 1 - 8e(0)(0) = 1 \][/tex]

[tex]\[ \frac{\partial F}{\partial y} = 1 - 8e(0)(0) = 1 \][/tex]

[tex]\[ \frac{\partial F}{\partial z} = 1 - 8e(0)(0) = 1 \][/tex]

So, the gradient vector of [tex]\(F\) at \((0, 0, 8)\) is \( \nabla F = (1, 1, 1) \).[/tex]

Therefore, the equation of the tangent plane is:

[tex]\[ x + y + z = 8 \][/tex]

(b) To find the equation of the normal line, we use the same gradient vector [tex]\( \nabla F = (1, 1, 1) \)[/tex] as the direction vector for the line.

Given the point \((0, 0, 8)\) on the surface, the parametric equations of the normal line are:

[tex]\[ x(t) = 0 + t(1) = t \][/tex]

[tex]\[ y(t) = 0 + t(1) = t \][/tex]

[tex]\[ z(t) = 8 + t(1) = 8 + t \][/tex]

So, the parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]

Imagine you conducted a study to look at the association between whether an expectant mother eats breakfast (or not) and the gender of her baby. Cramér's V = .22. How would you interpret this value? (Hint: Cramér's V can range between 0 and 1.) There was a small to medium association between baby gender and whether the mother ate breakfast every day. 22% of the variation in frequency counts of baby gender (boy or girl) can be explained by whether or not the mother ate breakfast every day. 2.2% of the variation in frequency counts of baby gender (boy or girl) can be explained by whether or not the mother ate breakfast every day. There is a medium to large association between the gender of the baby and whether or not the mother ate breakfast every day.

Answers

Answer:

: The percentage of newborn infants who are exclusively breastfed at the time of hospital discharge.

Data Source: Rhode Island Department of Health, Center for Health and Data Analysis, Newborn Developmental Risk Screening Program Database and Maternal and Child Health Database.

Footnotes: The year indicated is the mid-year of a five-year period of data.

Step-by-step explanation:

Final answer:

Cramér's V value of .22 in the study indicates a small to medium association between whether the mother ate breakfast and the gender of her baby.

Explanation:

In the study, Cramér's V measuress the strength of association between two nominal variables, in this case, the mother's breakfast-eating habits and the gender of her baby. The value of Cramer’s V ranges from 0 (indicating no association) to 1 (indicating a perfect association)). Given the value of Cramér's V at .22 for this study, there is a small to medium association between whether a mother ate breakfast and the gender of her baby. It does not mean that 22% of the variation in baby gender can be explained by the mother's breakfast-eating habits. Rather, it quantifies the extent of the association between these two factors.

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An archer is able to hit the bull's-eye 53% of the time. If she shoots 10 arrows, what is the probability that she gets exactly 4 bull's-eyes?


Answers

Final answer:

To find the probability that the archer gets exactly 4 bull's-eyes out of 10 shots, use the binomial probability formula. The probability is approximately 0.221.

Explanation:

To find the probability that the archer gets exactly 4 bull's-eyes out of 10 shots, we can use the binomial probability formula. The formula is:

P(x) = nCx * p^x * (1-p)^(n-x)

Where:

P(x) is the probability of getting exactly x successesn is the number of trialsx is the number of successes we want to find the probability forp is the probability of a single success

In this case, n = 10, x = 4, and p = 0.53 (since the archer hits the bull's-eye 53% of the time).

Plugging these values into the formula:

P(4) = 10C4 * (0.53)^4 * (1-0.53)^(10-4)

Calculating this gives us P(4) ≈ 0.221

Final answer:

To find the probability of exactly 4 bull's-eyes out of 10 shots for an archer with a 53% success rate, use the binomial probability formula, resulting in approximately a 20.6% chance.

Explanation:

The probability that an archer who hits the bull's-eye 53% of the time getting exactly 4 bull's-eyes out of 10 shots can be calculated using the binomial probability formula: P(X = k) = nCk × p^k × (1 - p)^(n - k), where:

n is the number of trials (10 arrows)

k is the number of successes (4 bull's-eyes)

p is the probability of success on a single trial (0.53)

nCk is the number of combinations of n things taken k at a time

Applying these values:

Calculate the number of combinations for getting 4 successes out of 10 trials:

10C4 = 210.

Calculate the probability of getting exactly 4 bull's-eyes:

P(X = 4) = 210 × (0.53)^4 × (0.47)^6.

Compute the result:

P(X = 4) ≈ 0.206 (rounded to three decimal places).

The archer has approximately a 20.6% chance of hitting exactly 4 bull's-eyes out of 10 shots.

Thomas purchased a used vehicle that depreciates under a straight-line
method. The initial value of the car is $7000, and the salvage value is $500. If
the car is expected to have a useful life of another 5 years, how much will it
depreciate each year?
O
A. $1300
O B. $1000
O c. $7000
O D. $5000

Answers

Answer: A. $1300

Step-by-step explanation:

The initial value is $7000

The Salvage value is $500

The car will depreciate under a straight-line method, meaning it will lose $6500 value in 5 years.

6500 / 5 = 1300

The diameters of a batch of ball bearings are known to follow a normal distribution with a mean 4.0 in and a standard deviation of 0.15 in. If a ball bearing is chosen randomly, find the probability of realizing the following event: (a) a diameter between 3.8 in and 4.3 in, (b) a diameter smaller than 3 9 in, (c) a diameter larger than 4.2 in

Answers

Answer:

a) 88.54% probability of a diameter between 3.8 in and 4.3 in

b) 25.14% probability of a diameter smaller than 3.9in

c) 90.82% probability of a diameter larger than 4.2 in

Step-by-step explanation:

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

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

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

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

In this problem, we have that:

[tex]\mu = 4, \sigma = 0.15[/tex]

(a) a diameter between 3.8 in and 4.3 in,

This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.

X = 4.3

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

[tex]Z = \frac{4.3 - 4}{0.15}[/tex]

[tex]Z = 2[/tex]

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

X = 3.8

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

[tex]Z = \frac{3.8 - 4}{0.15}[/tex]

[tex]Z = -1.33[/tex]

[tex]Z = -1.33[/tex] has a pvalue of 0.0918

0.9772 - 0.0918 = 0.8854

88.54% probability of a diameter between 3.8 in and 4.3 in

(b) a diameter smaller than 3 9 in,

This is the pvalue of Z when X = 3.9. So

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

[tex]Z = \frac{3.9 - 4}{0.15}[/tex]

[tex]Z = -0.67[/tex]

[tex]Z = -0.67[/tex] has a pvalue of 0.2514

25.14% probability of a diameter smaller than 3.9in

(c) a diameter larger than 4.2 in

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

[tex]Z = \frac{4.2 - 4}{0.15}[/tex]

[tex]Z = 1.33[/tex]

[tex]Z = 1.33[/tex] has a pvalue of 0.9082

90.82% probability of a diameter larger than 4.2 in

Hi to whoever is reading this. I really need help on breaking down the expression. I am doing this without a calculator and I don’t Know what steps to take to get the answer.

Answers

Answer:

A

Step-by-step explanation:

[tex]81[/tex] is [tex]9^2[/tex] which happens to be [tex](3^2)^2[/tex] or [tex]3^4[/tex].

So [tex]\sqrt[4]{81} =3[/tex]

this eliminates options C and D

It can't be option B as both b and c aren't 4th rooted.

49w = −8
what is the value of w

Answers

Answer:

w= -0.163265306122449

Step-by-step explanation:

49w = -8

divide both sides by 49

w= -8/49 = -0.163265306122449

check answer by multiplying -0.163265306122449  by 49

49 * -0.163265306122449 = -8

Answer:

w = -8/49

Step-by-step explanation:

49w = −8

Divide each side by 49

49w/49 = -8/49

w = -8/49

A recipe requires 3/4 cups of sugar for every 14 cookies how many cups of sugar will be needed to bake 168 cookies

Answers

Answer:

9 cups

Step-by-step explanation:

3/4 cups of sugar for 14 cookies

There are 168÷14 = 12 set of 14 cookies in 168

So we need 12 times 3/4 sugar

12 × 3/4 = 9 cups of sugar needed

1. Walking to improve health. In a study investigating a link between walking and improved health (Social Science & Medicine, Apr. 2014), researchers reported that adults walked an average of 5.5 days in the past month for the purpose of health or recreation. a) Specify the null and alternative hypotheses for testing whether the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days. b) Is the test one-tailed or two-tailed? c) Using α = .01, specify the critical value (CV), the rejection region, and draw the distribution.

Answers

Answer:

a) We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 5.5[/tex]  

Alternative hypothesis:[tex]\mu < 5.5[/tex]  

b) For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

c) For this case the significance level is given [tex]\alpha=0.01[/tex] and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

[tex]z_{cric}= -2.326[/tex]

And the rejection zone would be: [tex] (\infty ,-2.326)[/tex]

And the region is on the figure attached

Step-by-step explanation:

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 5.5[/tex]  

Alternative hypothesis:[tex]\mu < 5.5[/tex]  

Part b

For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

Part c

For this case the significance level is given [tex]\alpha=0.01[/tex]and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

[tex]z_{cric}= -2.326[/tex]

And the rejection zone would be: [tex] (\infty ,-2.326)[/tex]

And the region is on the figure attached

A deck of cards contains 52 cards.two decks are put together and divided among eight players.how many cards does each person get

Answers

Answer:

13 cards

Step-by-step explanation:

One deck has 52 cards.

2 decks have 2 * 52 cards, or 104 cards.

104 cards are divided among 8 players.

104/8 = 13

Answer: 13 cards

If Sergio has 16 quarters and dimes in his pocket, and they have a combined value of 265 cents, how many of each coin does he have?

Answers

9 dimes and 7 quarters. Each quarter is 25 cents and each dime is 10 cents. 25 x 7 = 175, 10 x 9 = 90. Add 175 and 90 together and it gives you 265.

State police believe that 70% of the drivers traveling on a major interstate highway exceed the speed limit. They plan to set up a radar trap and check the speeds of 80 cars. Taking a random sample of eighty cars is considered sufficiently large for the distribution of the sample proportion to be normally distributed. Would a random sample of 50 cars have been large enough and why?

Answers

Answer:

With n = 50, both conditions(np > 5, n(1-p) > 5) are satisfied, so a random sample of 50 cars would have been large enough.

Step-by-step explanation:

Binomial probability distribution

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

Can be approximated to the normal distribution if:

np > 5, n(1-p) > 5

70% of the drivers traveling on a major interstate highway exceed the speed limit.

So p = 0.7.

Would a random sample of 50 cars have been large enough and why?

With n = 50

np = 50*0.7 = 35 > 5

n(1-p) = 50*0.3 = 15 > 5

With n = 50, both conditions(np > 5, n(1-p) > 5) are satisfied, so a random sample of 50 cars would have been large enough.

sara pays $4100 tuition and fees for each of the two semesters, plus an additional $41" for textbooks each semester. what is the monthly cost?

Answers

Answer:

The monthly cost is $690.17.

Step-by-step explanation:

Each semester costs $4100.

In each semester, there are $41 spent on textbooks.

In a year:

2*4100 + 2*41 = $8282

Monthly cost

A year has 12 months. So

8282/12 = 690.17.

The monthly cost is $690.17.


The graph of a line passes through the points (0,5) and (-10,0). What is the equation
of the line?
CLEAR
CHECK
y = 2x - 10
y = 5x + 5
y= 1 / 8x 10
y = -2x + 5

Answers

Answer:

Y=1/2x +5

Step-by-step explanation:

(0,5) (-10,0)

0-5/-10-0=1/2

(0,5) (x, y)

Y-5/x-10=1/2(crossmultiply)

2y-10=x

2y= x +10(divide all sides by 2)

Y=1/2x+5

A bag of marbles contains 3 yellow, 6 blue, 1 green, 8 red, 2 orange. Find the probability: P(yellow, then orange WITHOUT replacement)

Answers

Answer:

P(yellow, then orange WITHOUT replacement) = 0.0158

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

This problem asks:

The probability of first getting a yellow marble, then an orange marble, without replacemtn.

Yellow marble first:

3 yellow marbles out of 3+6+1+8+2 = 20 marbles.

So P(a) = 3/20.

Then the orange marble:

Now the yellow marble is retired.

2 orange marbles out of 19.

So P(b) = 2/19

Probability:

P(yellow, then orange WITHOUT replacement) = P(a)P(b) = (3/20)*(2/19) = 0.0158

Find the sum of place value and face value of 7 in 1879523​

Answers

Answer:

Ok. so the Place value is 70000 and the face value is 7. Face value of a digit in a number is the digit itself.

More clearly, face value of a digit always remains same irrespective of the position where it is located

Place value of a digit in a number is the digit multiplied by thousand or hundred or whatever place it is situated.

Hope this helped.

Step-by-step explanation:

The trucking company has a policy that any truck with a mileage more than 2.5 standard deviations above the mean should be removed from the road and inspected. What is the probability that a randomly selected truck from the fleet will have to be inspected? ROUND YOUR ANSWER TO 4 DECIMAL PLACES

Answers

Answer:

0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected

Step-by-step explanation:

Z-score

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

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

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

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

What is the probability that a randomly selected truck from the fleet will have to be inspected?

Values of Z above 2.5. So this probability is 1 subtracted by the pvalue of Z = 2.5.

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected

Answer:

For this case we want this probability:

[tex] P(X > \mu +2.5 \sigma) [/tex]

And we can use the z score formula given by:

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

And replacing we got:

[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]

We want this probability:

[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]

And using the normal standard table or excel we got:

[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/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 mileage of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

For this case we want this probability:

[tex] P(X > \mu +2.5 \sigma) [/tex]

And we can use the z score formula given by:

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

And replacing we got:

[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]

We want this probability:

[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]

And using the normal standard table or excel we got:

[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]

How do the areas of triangle ABC and DEF compare?
O The area of ABC is 1 square unit less than the area of
ADEF
O The area of AABC is equal to the area of ADEF.
O The area of ABC is 1 square unit greater than the
area of ADEF.
O The area of ABC is 2 square units greater than the
area of ADEF.

Answers

Answer:

C: The area of ABC is 1 square unit greater than the

area of ADEF

Step-by-step explanation:

Due to the area of triangle ABC and triangle DEF, the ABC triangle has dimensions that are 1 square unit greater that the area of ADEF. This is because the lines have different lengths. Good Luck on the Rest of Your Questions!!

Answer: C

Step-by-step explanation:

Cuál es el valor de la expresión -2+3(8+-13)? ​

Answers

Answer:

-17

Step-by-step explanation:

-2+3(8+-13)

-2+3(8-13)

-2+3(-5)

-2-15

-17

[tex]\text{Solve:}\\\\-2+3(8+-13)\\\\\text{Solve the variables inside of the parenthesis}\\\\-2+3(-5)\\\\-2-15\\\\\boxed{-17}[/tex]

Tiffany sells cars. She gets an 18% commission. How much will she make on a $ 28,000.00 car?

Answers

Answer:

5040$

Step-by-step explanation:

28000 x .18 = 5040

We are given that if Tiffany closes any car on a car she gets 18% as commission. For a car that cost $28,000 her commission will be  $5,040

Given data

Commission = 18%

Cost of car  =  $ 28,000.00

Let us find 18% of  $28,000.00

Commission = 18/100* 28,000.00

Commission = 0.18* 28,000.00

Commission = $5,040

Hence her commission will be: $5,040

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Find the derivative.
StartFraction d Over dt EndFraction Integral from 0 to t Superscript 8 StartRoot u cubed EndRoot du
a. by evaluating the integral and differentiating the result.
b. by differentiating the integral directly.

Answers

Answer:

Using either method, we obtain:  [tex]t^\frac{3}{8}[/tex]

Step-by-step explanation:

a) By evaluating the integral:

 [tex]\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du[/tex]

The integral itself can be evaluated by writing the root and exponent of the variable u as:   [tex]\sqrt[8]{u^3} =u^{\frac{3}{8}[/tex]

Then, an antiderivative of this is: [tex]\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}[/tex]

which evaluated between the limits of integration gives:

[tex]\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}[/tex]

and now the derivative of this expression with respect to "t" is:

[tex]\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}[/tex]

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

[tex]g(x)=\int\limits^x_a {f(t)} \, dt[/tex]

is continuous on [a,b], differentiable on (a,b) and  [tex]g'(x)=f(x)[/tex]

Since this this function [tex]u^{\frac{3}{8}[/tex] is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

[tex]\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}[/tex]

A random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph. From past information, it is known that the standard deviation of the population is 22 mph. The 95 percent confidence interval for μ is determined as (61.08, 68.92). If we are to reduce the sample size to 100 (other factors remain unchanged), the 95 percent confidence interval for μ would: Question 6 options: be the same. become wider. become narrower. be impossible to determine.

Answers

Answer:

95 percent confidence interval for μ is determined by

(60.688 ,  69.312)

a) be the same

Step-by-step explanation:

Step (i):-

Given data a random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph.

The sample size is n= 121

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (ii) :-

Given  The 95 percent confidence interval for μ is determined as

(61.08, 68.92)

we are to reduce the sample size to 100 (other factors remain unchanged) so

given The sample size is n= 100

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (iii):-

95 percent confidence interval for μ is determined by

[tex](x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )[/tex]

[tex](65 - 1.96\frac{22}{\sqrt{100} } , 65 + 1.96 \frac{22}{\sqrt{100} } )[/tex]

(65 - 4.312 ,65 +4.312)

(60.688 ,  69.312) ≅(61 ,69)

This is similar to (61.08, 68.92) ≅(61 ,69)

Conclusion:-

it become same as (61.08, 68.92)

Final answer:

In the given scenario, the 95 percent confidence interval for μ would become wider if the sample size is reduced from 121 to 100. This is due to the increased uncertainty in estimation as a result of a smaller sample size.

Explanation:

In statistics, a confidence interval is a range in which a population parameter is estimated to lie. The calculation of a confidence interval involves several factors, one of which is the sample size. The width of the confidence interval decreases as the sample size increases, assuming that all other factors remain the same.

In this case, the sample size is proposed to decrease from 121 to 100. As a result, the confidence interval for μ would become wider. This is because as the sample size decreases, the standard error increases, which in turn widens the confidence interval, allowing for more uncertainty in estimations.

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A worker has asked her supervisor for a letter of recommendation for a new job. She estimates that there is an 80 percent chance that she will get the job if she receives a strong recommendation, a 40 percent chance if she receives a moderately good recommendation, and a 10 percent chance if she receives a weak recommendation. She further estimates that the probabilities that the recommendation will be strong, moderate, and weak are 0.7, 0.2, and 0.1, respectively.

a. How certain is she that she will receive the new job offer?
b. Given that she does receive the offer, how likely should she feel that she received a strong recommendation?

i. a moderate recommendation?
ii. a weak recommendation?

c. Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation?
i. a moderate recommendation?
ii. a weak recommendation?

Answers

Answer:

a) P(Job offer) = 0.65

b) P(S|J) = 0.862

P(M|J) = 0.143

P(W|J) = 0.0179

c) P(S|J') = 0.400

P(M|J') = 0.343

P(W|J') = 0.257

Step-by-step explanation:

- Let the event that she gets a job be J

- Let the event that she does not get the job be J'

- Let the event that she receives a strong recommendation be S

- Let the event that she receives a moderate recommendation be M

- Let the event that she receives a weak recommendation be W.

Given in the question,

P(J|S) = 80% = 0.8

P(J|M) = 40% = 0.4

P(J|W) = 10% = 0.1

P(S) = 0.70

P(M) = 0.20

P(W) = 0.10

a) How certain is she that she will receive the new job offer?

P(J) = P(J n S) + P(J n M) + P(J n W) (since S, M and W are all of the possible outcomes that lead to a job)

But note that the conditional probability, P(A|B) is given mathematically as,

P(A|B) = P(A n B) ÷ P(B)

P(A n B) is then given as

P(A n B) = P(A|B) × P(B)

So,

P(J n S) = P(J|S) × P(S) = 0.80 × 0.70 = 0.56

P(J n M) = P(J|M) × P(M) = 0.40 × 0.20 = 0.08

P(J n W) = P(J|W) × P(W) = 0.10 × 0.10 = 0.01

P(J) = P(J n S) + P(J n M) + P(J n W)

P(J) = 0.56 + 0.08 + 0.01 = 0.65

b) Given that she does receive the offer, how likely should she feel that she received a strong recommendation?

This probability = P(S|J)

P(S|J) = P(J n S) ÷ P(J) = 0.56 ÷ 0.65 = 0.862

i. a moderate recommendation?

P(M|J) = P(J n M) ÷ P(J) = 0.08 ÷ 0.65 = 0.143

ii. a weak recommendation?

P(W|J) = P(J n W) ÷ P(J) = 0.01 ÷ 0.65 = 0.0179

c) Probability that she doesn't get job offer, given she got a strong recommendation = P(J'|S)

P(J'|S) = 1 - P(J|S) = 1 - 0.80 = 0.20

Probability that she doesn't get job offer, given she got a moderate recommendation = P(J'|M)

P(J'|M) = 1 - P(J|M) = 1 - 0.40 = 0.60

Probability that she doesn't get job offer, given she got a weak recommendation = P(J'|S)

P(J'|W) = 1 - P(J|W) = 1 - 0.10 = 0.90

Total probability that she doesn't get job offer

P(J') = P(J' n S) + P(J' n M) + P(J' n W)

P(J' n S) = P(J'|S) × P(S) = 0.20 × 0.70 = 0.14

P(J' n M) = P(J'|M) × P(M) = 0.60 × 0.20 = 0.12

P(J' n W) = P(J'|W) × P(W) = 0.90 × 0.10 = 0.09

Total probability that she doesn't get job offer

P(J') = P(J' n S) + P(J' n M) + P(J' n W)

= 0.14 + 0.12 + 0.09 = 0.35

Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation?

This probability = P(S|J')

P(S|J') = P(J' n S) ÷ P(J') = 0.14 ÷ 0.35 = 0.400

i. a moderate recommendation?

P(M|J') = P(J' n M) ÷ P(J') = 0.12 ÷ 0.35 = 0.343

ii. a weak recommendation?

P(W|J') = P(J' n W) ÷ P(J') = 0.09 ÷ 0.35 = 0.257

Hope this Helps!!!

Final answer:

She has 62% certainty of getting the job. If she gets it, there are high chances (~90%) that the recommendation was strong. In case of rejection, she's slightly more likely to have received a strong recommendation than a moderate or weak one.

Explanation:

The subject of this problem involves the understanding of conditional probabilities and the principle of total probability. Let's denote the events as follows:

A1: the recommendation is strong A2: the recommendation is moderate A3: the recommendation is weak B: she gets the new job offer

a. The total probability that she will receive the job offer is calculated as follows:

P(B)=P(A1)P(B|A1)+P(A2)P(B|A2)+P(A3)P(B|A3) = 0.7*0.8+0.2*0.4+0.1*0.1=0.62, so she is 62% certain that she will receive the job offer.

b. If she does receive the offer, the probabilities that she got a strong, moderate and weak recommendation are calculated using Bayes' Theorem:

P(A1|B)=P(B|A1)P(A1)/P(B)=0.8*0.7/0.62 =~ 0.9 P(A2|B)=P(B|A2)P(A2)/P(B)=0.4*0.2/0.62 =~ 0.13 P(A3|B)=P(B|A3)P(A3)/P(B)=0.1*0.1/0.62 =~ 0.016

c. If she doesn't get the job, the probabilities are calculated similarly:

P(A1|Not B)=P(Not B|A1)P(A1)/P(Not B)=0.2*0.7/0.38 =~ 0.37 P(A2|Not B)=P(Not B|A2)P(A2)/P(Not B)=0.6*0.2/0.38 =~ 0.32 P(A3|Not B)=P(Not B|A3)P(A3)/P(Not B)=0.9*0.1/0.38 =~ 0.24

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The same researcher wants to see if recently widowed spouses leave the house more after they perform some activity (e.g., Bingo, Support Group, etc.). She wants to compare participants before and after a group activity. Which type of t-test should she use?

Answers

Answer:

paired t-test

Step-by-step explanation:

Paired t-test is used to measure before and after effect. For example, a researcher wants to check the effect of new teaching method and he wants to compares marks of students before new teaching method and marks of students before new teaching method.

In the given example there are two groups participants before group activity and participants after group activity. A researcher wants to investigate that whether recently widowed spouses leave house after performing group activity or not. The observations in these two groups are paired naturally and we take the differences of these two groups are considered as random sample. This type of t-test is known as paired t-test.

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