A large company that must hire a new president prepares a final list of five candidates, all of whom are equally qualified. Two of these candidates are members of a minority group. To avoid bias in the selection of the candidate, the company decides to select the president by lottery. a. What is the probability one of the minority candidates is hired

Answers

Answer 1

Final answer:

The probability of one of the minority candidates being hired is 40%.

Explanation:

To find the probability that one of the minority candidates is hired, we need to determine the number of favorable outcomes (one of the minority candidates being selected) and divide it by the total number of possible outcomes (selecting any candidate from the final list of five).

Since there are two minority candidates and five candidates total, the probability of selecting one of the minority candidates is

P(one of the minority candidates being hired) = 2/5 = 0.4 = 40%


Related Questions

Find the mean absolute deviation for the data set.
5,6,6,8,10

Answers

Answer: 1.6

Order the numbers

5,6,6,8,10

Add

5+6+6+8+10=35

Divide

35÷5=7

Mean: 7

Sum divided by the count.

Final Answer: 1.6

Answer:

1.6

Step-by-step explanation:

Mean: 5 + 6 + 6 + 8 + 10 = 35/5 = 7

7 - 5 = 2

7 - 6 = 1

7 - 6 = 1

7 - 8 = 1

7 - 10 = 3

1 + 1 + 1 + 2 + 3 = 8/5 = 1.6

round 12.566370614359 to the nearest hundredth

Answers

Answer:12.57

Step-by-step explanation:

Answer:

12.566370614359 rounded to the nearest hundreth is 12.57

The popular candy Skittles comes in 5 colors. According to the Skittles website, the 5 colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population. Suppose that we purchase a small bag of Skittles. Assume this size bag always has 40 candies. In this particular bag 10 are green. What is the probability that a randomly selected bag of this size has 10 or more green candies

Answers

Answer:

27.76% probability that a randomly selected bag of this size has 10 or more green candies

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 40, p = 0.2[/tex]

So

[tex]\mu = E(X) = np = 40*0.2 = 8[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{40*0.2*0.8} = 2.53[/tex]

What is the probability that a randomly selected bag of this size has 10 or more green candies

Using continuity correction, this is [tex]P(X \geq 10 - 0.5) = P(X \geq 9.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 9.5. So

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

[tex]Z = \frac{9.5 - 8}{2.53}[/tex]

[tex]Z = 0.59[/tex]

[tex]Z = 0.59[/tex] has a pvalue of 0.7224

1 - 0.7224 = 0.2776

27.76% probability that a randomly selected bag of this size has 10 or more green candies

Answer:

[tex]P(x\geq 10)=0.2682[/tex]

Step-by-step explanation:

The number x of green candies in a bag of 40 candies follows a binomial distribution, because we have:

n identical and independent events: 40 candiesa probability p of success and (1-p) of fail: a probability of 0.2 to get a green candie and 0.8 to doesn't get a green candie.

So, the probability that in a bag of 40 candies, x are green is calculated as:

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

Replacing, n by 40 and p by 0.2, we get:

[tex]P(x)=\frac{40!}{x!(40-x)!}*0.2^{x}*(1-0.2)^{40-x}[/tex]

So, the probability that a randomly selected bag of this size has 10 or more green candies is equal to:

[tex]P(x\geq 10)=P(10)+P(11)+...+P(40)\\P(x\geq 10)=1-P(x<10)[/tex]

Where [tex]P(x<10)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)[/tex]

So, we can calculated P(0) and P(1) as:

[tex]P(0)=\frac{40!}{0!(40-0)!}*0.2^{0}*(1-0.2)^{40-0}=0.00013\\P(1)=\frac{40!}{1!(40-1)!}*0.2^{1}*(1-0.2)^{40-1}=0.00133[/tex]

At the same way, we can calculated P(2), P(3), P(4), P(5), P(6), P(7), P(8) and P(9) and get that P(x<10) is equal to:

[tex]P(x<10)=0.7318[/tex]

Finally, the probability [tex]P(x\geq 10)[/tex] that a randomly selected bag of this size has 10 or more green candies is:

[tex]P(x\geq 10)=1-P(x<10)\\P(x\geq 10)=1-0.7318\\P(x\geq 10)=0.2682[/tex]

Consider the four numbers a, b, c, d with a ≤ b ≤ c ≤ d, where a, b, c, d are integers. The mean of the four numbers is 4.The mode is 3. The median is 3.The range is 6. Find d

Answers

Answer:

d = 2

Step-by-step explanation:

We have four unknown numbers a, b, c, d

It is given that the mode is 3,

Since the mode is 3 then at least two numbers are 3.

It is given that the median is 3,

Since the median is 3 which means the middle two values must be 3

a, 3, 3, d

It is given that the mean of the four numbers is 4,

Since the mean of the four number is 4 then

mean = (a + 3 + 3 + d)/4

4 = (a + 6 + d)/4

4*4 = a + 6 + d

16 = a + 6 + d    eq. 1

It is given that the range is 6,

Since the range is 6 which is the difference between highest and lowest number that is

a - d = 6

a = 6 + d    eq. 2

Substitute the eq. 2 into eq. 1

16 = a + 6 + d

16 = (6 + d) + 6 + d

16 = 12 + 2d

2d = 16 - 12

d = 4/2

d = 2

Substitute the value of d into eq. 2

a = 6 + d

a = 6 + 2

a = 8

so

a, b, c, d = 8, 3, 3, 2

Verification:

a ≤ b ≤ c ≤ d

8 ≤ 3 ≤ 3 ≤ 2

mean = (a + b + c + d)/4

mean = (8 + 3 + 3 + 2)/4

mean = 16/4

mean = 4

range = a - d

range = 8 - 2

range = 6

A car rental agency rents 440 cars per day at a rate of ​$30 per day. For each ​$1 increase in​ rate, 10 fewer cars are rented. At what rate should the cars be rented to produce the maximum​ income? What is the maximum​ income

Answers

The cars should be rented at a rate of $30 per day to produce the maximum income of $13200 per day.

To solve this problem, we can use the following steps:

Define the variables.

Let x be the number of cars rented per day and y be the rate per day.

Write down the equation for the revenue.

The revenue is equal to the number of cars rented multiplied by the rate per day. Therefore, the equation for the revenue is:

R = x * y

Write down the equation for the decrease in demand. For each $1 increase in rate, 10 fewer cars are rented. Therefore, the equation for the decrease in demand is:

D = -10 * (y - 30)

Set the revenue equal to the maximum value. The revenue is maximized when the derivative of the revenue function is zero. Therefore, we set the derivative of the revenue function equal to zero and solve for y.

R' = x = 0

x = 440

y = 30

Calculate the maximum revenue. The maximum revenue is equal to the number of cars rented multiplied by the rate per day. Therefore, the maximum revenue is:

R = 440 * 30 = $13200

Therefore, the cars should be rented at a rate of $30 per day to produce the maximum income of $13200 per day.

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Let p equal the proportion of drivers who use a seat belt in a country that does not have a mandatory seat belt law. It was claimed that p = 0.14. An advertising campaign was conducted to increase this proportion. Two months after the campaign, y = 104 out of a random sample of n = 590 drivers were wearing their seat belts . Was the campaign successful? (a) Define the null and alternative hypotheses. (b) Define a rejection region with an α = 0.01 significance level. (c) Determine the approximate p-value and state your conclusion.

Answers

Answer:

a)Null hypothesis:[tex]p\leq 0.14[/tex]  

Alternative hypothesis:[tex]p > 0.14[/tex]  

b) For this case we are conducting a right tailed test so then we need to look in the normal standard distribution a quantile that accumulates 0.01 of the are in the left and we got;

[tex]z_{crit} = 2.33[/tex]

So then the rejection region would be [tex](2.33 , \infty)[/tex]

c) The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.52)=0.006[/tex]  

And since the p value is lower than the significance level then we can reject the null hypothesis. So then we can conclude that the true proportion of interest is higher than 0.14 at 1% of significance.

Step-by-step explanation:

Data given and notation

n=590 represent the random sample taken

X=104 represent the drivers were wearing their seat belts

We can estimate the sample proportion like this:

[tex]\hat p=\frac{104}{590}=0.176[/tex] estimated proportion of  drivers were wearing their seat belts

[tex]p_o=0.14[/tex] is the value that we want to test

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

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

a) System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion of drivers were wearing their seat belts is higher than 0.14 or no, so the system of hypothesis are.:  

Null hypothesis:[tex]p\leq 0.14[/tex]  

Alternative hypothesis:[tex]p > 0.14[/tex]  

Part b

For this case we are conducting a right tailed test so then we need to look in the normal standard distribution a quantile that accumulates 0.01 of the are in the left and we got;

[tex]z_{crit} = 2.33[/tex]

So then the rejection region would be [tex](2.33 , \infty)[/tex]

Part c

The statistic is given by:

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.176 -0.14}{\sqrt{\frac{0.14(1-0.14)}{590}}}=2.52[/tex]  

Statistical decision  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.52)=0.006[/tex]  

And since the p value is lower than the significance level then we can reject the null hypothesis. So then we can conclude that the true proportion of interest is higher than 0.14 at 1% of significance.

Answer:

See explanation

Step-by-step explanation:

Solution:-

- Let the proportion of drivers who use a seat belt in a country that does not have a mandatory seat belt law = p.

- A claim was made that p = 0.14. We will state our hypothesis:

                Null hypothesis: p = 0.14

- Two months after the campaign, y = 104 out of a random sample of n = 590 drivers were wearing their seat belts. The sample proportion can be determined as:

               Sample proportion ( p^ ) = y / n = 104 / 590 = 0.176

- The alternate hypothesis will be defined by a population proportion that supports an increase. So we state the hypothesis:

               Alternate hypothesis: p > 0.14

- The rejection is defined by the significance level ( α = 0.01 ). The rejection region is defined by upper tail of standard normal.

- The Z-critical value that limits the rejection region is defined as:

                            P ( Z < Z-critical ) = 1 - 0.01 = 0.99

                            Z-critical = 2.33

- All values over Z-critical are rejected.

- Determine the test statistics by first determining the population standard deviation ( σ ):

- Estimate σ using the given formula:

                      σ = [tex]\sqrt{\frac{p*(1-p)}{n} } = \sqrt{\frac{0.14*(1-0.14)}{590} }= 0.01428[/tex]

- The Z-test statistics is now evaluated:

                     Z-test = ( p^ - p ) / σ

                     Z-test = ( 0.1763 - 0.14 ) / 0.01428

                    Z-test = 2.542

- The Z-test is compared whether it lies in the list of values from rejection region.

                     2.542 > 2.33

                     Z-test > Z-critical

Hence,

                     Null hypothesis is rejected

- The claim made over the effectiveness of campaign is statistically correct.

Suppose, Gallup asks 2100 Japanese people whether Japan will be able to completely recover from the devastation of the recent earthquake/Tsunami and and that 67% believed in the affirmative. Based on the margin of error, what should be the population percentage of Japanese who believes in the complete recovery of Japan

Answers

Answer:

Option E) The population percentage of Japanese who believes in the complete recovery of Japan is Between 64.82% and 69.18%.

Step-by-step explanation:

Gallup asked 2100 Japanese, so the sample size is:

n = 2100

67% of the Japanese answered in affirmative. This means the proportion of population which answered in favor or affirmative is:

p = 67%

Based on his findings, Gallup constructed a confidence interval. We have to identify the correct confidence interval i.e. the population percentage of Japanese who believes in the complete recovery of Japan.

The confidence interval will always be in form of a range of values i.e. between two values: A lower limit and an upper limit. This automatically removes choices A and B from the list of correct answers.

Furthermore, the confidence interval is symmetric about the sample proportion(p), as the formula to calculate the confidence interval for a population proportion is:

( p - M.E, p + M.E )

where M.E means Margin of Error. Since, same value(M.E) is added to and subtracted from the sample proportion(p), the confidence interval will be symmetric about the sample proportion.

So, now we will find if the values in choices C,D and E are symmetric about the mean or not. If the values are symmetric the difference of the values in each option from p = 67% must be same.

Choice C)

60% and 70%

We can easily tell that these values are not symmetric about 67%. Therefore, this cannot be the answer.

Choice D)

65.13% and 70.21%

67% - 65.13% = 1.87%

70.21% - 67% = 3.21%

These two values are not symmetric either. So these cannot be our confidence interval.

Choice E)

64.82% and 69.18%

67% - 64.82% = 2.18%

69.18% - 67% = 2.18%

These two values are same distance apart from 67%, this means they are symmetric about the sample proportion. Hence, choice E is the correct confidence interval. The Margin of Error is 2.18%

The population percentage of Japanese who believes in the complete recovery of Japan is Between 64.82% and 69.18%.

Plz help me with my homework

Answers

Answer:

Option D, 72 cubic inches

Step-by-step explanation:

The formula for the volume of a rectangular prism is length*width*height, which in this case is 3*3*8=72 cubic inches, or option D. Hope this helps!

Answer: D) 72

Step-by-step explanation: To get the volume of the rectangular prism all you got do is multiply the width x length x height. Therefore:

3 x 3 x 8 = 72

The answer is D) 72

(Hope this helps)

drag the correct step into order to evaluate 27- t x 3 for t = 6

Answers

Use PEMDAS and do the multiplication first.

Answer:

All steps below

Step-by-step explanation:

27 - t × 3

27 - 3t

t = 6

27 - 3(6)

27 - 18

9

The table represents a linear function. Find the values of a, b, and c. Show your work.
x | y
a 7
3 8
5 9
7 b
c 11

Answers

Answer:

a = 1, b = 10, c = 9

Step-by-step explanation:

x is increasing by positive two from each number and y is increasing by 1 from each number.

mx+n=y

m*3+n=8

m*5+n=9

=>m*5+n-(m*3+n)=9-8

      2m=1  => m=1/2

½ *3+ n =8  

n=8-3/2

n=13/2

=> y=x/2 +13/2

y=(x+13)/2

7=(a+13)/2; a+13=14  => a=1

b=(7+13)/2; b=20/2; b=10

11=(c+13)/2, c+13=22 => c=9

Guido is a citizen and resident of Belgium. He has a full-time job in Belgium and has lived there with his family for the past 10 years. In 2017, Guido came to the United States for the first time. The sole purpose of his trip was business. He intended to stay in the United States for only 180 days, but he ended up staying for 200 days because of unforeseen problems with his business. Guido came to the United States again on business in 2018 and stayed for 180 days. In 2019 he came back to the United States on business and stayed for 70 days. Under the substantial presence test defining a resident alien, how many days was Guido present in the U.S. in 2018

Answers

Answer:

Guido stayed in US in 2018 for 180 days which are greater than 31 days.

Guido Stayed in US in 2018 and in 2017 = (180 + 66) > 183 days.

So, yes Guido does meet US Statutory definition in 2018 and stayed 180 days in 2018.  

Step-by-step explanation:

Let's find out how many days in total Guido stayed in US in these 3 years 2017, 2018 and 2019.

Year = 2017

Days = 200

Year = 2018

Days = 180

Year = 2019

Days = 70

Total days stayed = 200 + 180 + 70

Total Days Stayed = 450 days.

In U.S, there are two tests are in place and for non-citizen of U.S and for the resident Alien or non - resident Alien status, one must pass one of these two tests, which are as follows:

1. Green Card Test:

2. Substantial Presence Test:

Here, in this problem, Guido is citizen and resident of Belgium. So, will check his criteria according to the substantial presence test.

So, the question is: how many days Guido was present in the U.S in 2018 under resident alien status.

In 2018, Guido stayed in US for 180 days.

So, according to the Substantial Presence test, one must be physically present in US for more than 31 days to be eligible for resident alien status. In addition, in 2018, his total of physical presence in US in 2018 and one third of physical presence in 2017 must be greater than 183 days.

If we see, both conditions are matched in case of Guido.

Guido stayed in US in 2018 for 180 days which are greater than 31 days.

Guido Stayed in US in 2018 and in 2017 = (180 + 66) > 183 days.

So, yes Guido does meet US Statutory definition in 2018 and stayed 180 days in 2018.  

Final answer:

In 2018, Guido was present in the United States for a total of 180 days according to the substantial presence test for determining resident alien status for tax purposes.

Explanation:

Under the substantial presence test, which is used to determine if an individual is a resident alien for tax purposes in the United States, only the days present in the current year (which, in the scenario provided, is 2018) are counted in full. In this case, Guido was present in the U.S. for 180 days in 2018. Days from prior years are counted partially, with only 1/3 of the days in the first preceding year and 1/6 of the days from the second preceding year included in the calculation. However, when determining how many days were present in a given year, such as 2018 in this example, only the days in that specific year are relevant. Therefore, according to the substantial presence test, in 2018, Guido was present in the United States for a total of 180 days.

Researchers are studying two populations of wild horses living in the western regions of a country. In a random sample of 32 horses taken from the first population, the mean age of the sample was 21 years. In a random sample of 41 horses from the second population, the mean age of the sample was 19 years. Is the sampling distribution of the difference in sample mean ages approximately normal?

A Yes, because the two populations of wild horses can be modeled by a normal distribution.
B Yes, because the samples were selected at random.
C Yes, because the sample sizes are both greater than 30.
D No, because the populations are not normal.
E No, because the difference in sample mean ages was not 0.

Answers

Answer:

Correct option: (C) Yes, because the sample sizes are both greater than 30.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

[tex]\mu_{\bar x}=\mu[/tex]

And the standard deviation of the distribution of sample mean is given by,

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

For the first sample, the sample size of the sample selected is:

n₁ = 32 > 30

Ans for the second sample, the sample size of the sample selected is:

n₂ = 41 > 30

Both the samples selected are quite large.

So, the Central limit theorem can be used to approximate the distribution of of the two sample means.

Ans since the distribution of the two sample means follows a normal distribution, the difference of the two means will also follows normal distribution.

Thus, the correct option is (C).

C Yes, because the sample sizes are both greater than 30.

The following information should be considered;

Given that, [tex]n_1 = 32[/tex] and [tex]n_2 = 41[/tex]Here both sample size should be more than 30.By applying the central limit theorem, sampling distribution of difference should be normal. Therefore, the third option is correct.

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Geometry question, Major points!! Please help

Answers

Answer:

C. x = 3cm

Step-by-step explanation:

The formula for volume of a triangular prism is

V = base * height

x in this case equals height, and the base is the area of the triangular base, which is

A(triangle) = 1/2 b*h

                 = 1/2 7*10 = 35 square centimeters

Plug this back into our formula:

V = base * height

105 = 35x

Solve for x.

105/35 = x

C. x = 3 cm

Answer:

C. 3 cm

Step-by-step explanation:

The volume of a triangular prism is denoted by: [tex]V=Bh[/tex], where B is the base area and h is the height.

Here, the base is actually the triangle, and we can calculate this area by using the formula for a triangle's area: [tex]A=\frac{1}{2} bh[/tex]. Here, b = 10 and h = 7, so:

[tex]A=\frac{1}{2} bh[/tex]

[tex]A=\frac{1}{2} *10*7=5*7=35[/tex] cm squared

Now, the height of the prism is x and we already know the volume is 105, so plug these values in:

[tex]V=Bh[/tex]

[tex]105=35*x[/tex]

x = 105/35 = 3

The answer is C.

Find the circumference for the given situation below. Round your answer to the
nearest tenth. Use 3.14 for pi.
The world's tallest Ferris wheel is in Osaka, Japan, and stands 369 feet tall. Its wheel
has a diameter of 328 feet. Find circumference of the Ferris wheel.​

Answers

Answer:

C = 1029.92 ft

Step-by-step explanation:

C = πd              Use the equation for circumference

C = 3.14(328)       Multiply

C = 1029.92 ft

If this answer is correct, please make me Brainliest!

Dr. Pagels is a mammalogist who studies meadow and common voles. He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices. So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles.

Define the null hypothesis.

Answers

Answer:

There is no relationship between voles and food preference (The food preferences among vole species are independent of one another)

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. The null hypothesis tries to establish that the old theory is true. So, for the case above, the old theory is that there is no relationship between voles species and food preference which is the null hypothesis.

The null hypothesis in Dr. Pagels' study is that there is no difference in the food preferences for peanut butter-oatmeal mixture and apple slices between meadow and common voles.

In the study conducted by Dr. Pagels for determining the food preference of meadow and common voles, the null hypothesis posits that there is no difference in preference for the peanut butter-oatmeal mixture or apple slices between the two species of voles. This implies that both meadow and common voles would choose either food option at an equal rate, suggesting that any observed preference could be attributed to random chance rather than a true preference.

The purpose of the null hypothesis is to establish a baseline expectation that no effect or difference is present, which can then be challenged by the experimental data. If significant differences in the vole populations' food choices are observed, the null hypothesis may be rejected, pointing to a potential preference for one food type over the other among the different vole species.

A. 7/25
B. 24/25
C. 7/24
D. 24/7

Answers

Answer:

D. 24/7

Step-by-step explanation:

SOH CAH TOA

we doing the tangent so

tan (α) = opposite / adjacent

tan (α) = 24/7

PLEASE HELP!!!


Explain the difference between P(A|B) and P(A)and P(B) given that events A and B are independent events.

Answers

Answer:

  for independent A and B, P(A|B) = P(A)

Step-by-step explanation:

The definition of  conditional probability is ...

  P(A|B) = P(A&B)/P(B)

When A and B are independent, ...

  P(A&B) = P(A)·P(B)

so the conditional probability is ...

  P(A|B) = (P(A)·P(B))/P(B) = P(A) . . . . . for independent A and B

In words, when A and B are independent, the probability of A given B is the same as the probability of A. That is, the probability of B has no effect on the probability of A.

Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly selected individual has the disease. Suppose n individuals are independently selected for testing. One way to proceed is to carry out a separate test on each of the n blood samples. A potentially more economical approach, group testing, was introduced during World War II to identify syphilitic men among army inductees. First, take a part of each blood sample, combine these specimens, and carry out a single test. If no one has the disease, the result will be negative, and only the one test is required. If at least one individual is diseased, the test on the combined sample will yield a positive result, in which case the n individual tests are then carried out. [The article "Random Multiple-Access Communication and Group Testing"† applied these ideas to a communication system in which the dichotomy was active/idle user rather than diseased/nondiseased.] If p = 0.15 and n = 5, what is the expected number of tests using this procedure? (Round your answer to three decimal places.)

Answers

Final answer:

The expected number of tests using the group testing procedure is 1.75.

Explanation:

To find the expected number of tests using the group testing procedure, we need to consider the different possible outcomes. Let's break it down:

If no one has the disease (probability = 1 - p), then only one test is required.

If at least one individual has the disease (probability = p), the test on the combined sample will yield a positive result, and then the n individual tests will be carried out.

Therefore, the expected number of tests is:

Expected number of tests = (probability of no disease) * (number of tests in this case) + (probability of disease) * (number of tests in this case)

For the first case, the number of tests is 1.

For the second case, the number of tests is n + 1, because one additional test is required after the positive result from the combined sample.

Substituting the values, Expected number of tests = (1 - p) * 1 + p * (n + 1)

Given p = 0.15 and n = 5, substituting the values we get:

Expected number of tests = (1 - 0.15) * 1 + 0.15 * (5 + 1) = 0.85 * 1 + 0.15 * 6 = 0.85 + 0.9 = 1.75

Therefore, the expected number of tests using this procedure is 1.75.

1/2(8x-39) = 1/4(12x + 32)

Answers

Answer:

please like and Mark as brainliest

don't mind my messy handwriting

the explanation is in the picture

Determine the perimeter of angle AGN using the picture.

Answers

Answer:

150

Step-by-step explanation:

Two tangents to the circle (until the point where the tangent and circle touch) has the same length, therefore, AR=AT, RG=GE, EN=NT

Then just add up everything!

A clock has a minute hand that is 6 in. long. How far has the tip of the minute hand traveled between 10:25 a.m. and 11:00 a.m.? Round your answer to the nearest tenth of an inch.

Answers

Answer:

Approximately = 20.0 inch to the nearest tenth

Step-by-step explanation:

We are going to calculate this by using the formula of the circle in a way.

Radius = 6 inches

Now between 10:25 a.m. and 11:00 a.m , the minute hand moved 35 minutes.

But there are total of 60 minutes in the clock which makes it a complete circle.

So 60 minutes = 2π

35 minutes = ?.

35 minutes =( 35*2π)/60

35 minutes = 1.166667π

So the distance covered by the minute hand = 1.166667π * 6 inches

= 21.991 inches

Approximately = 20.0 inch to the nearest tenth



please help brainly crown and 5-stars⭐️


Which expression is equivalent to 625 in exponential form?

A) 54

B) 53

C) 25 × 25

D) 5 × 125

Answers

Answer:

choice a....625 = [tex]5^{4}[/tex] in exponential form

Step-by-step explanation:

625 = 25*25 =  [tex]25^{2}[/tex]  =  [tex]5^{4}[/tex]

Answer:

choice A well be correct !! (did it on usa test prep)

Step-by-step explanation:

At a local university, a sample of 49 evening students was selected in order to determine whether the average age of the evening students is significantly different from 21. The average age of the students in the sample was 23 years. The population standard deviation is known to be 3.5 years. Determine whether or not the average age of the evening students is significantly different from 21. Use a 0.1 level of significance.

Answers

Answer:

[tex]z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4[/tex]    

[tex]p_v =2*P(z>4)=0.0000633[/tex]  

When we compare the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can reject the null hypothesis at 10% of significance. So the  the true mean is difference from 21 at this significance level.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=23[/tex] represent the sample mean

[tex]\sigma=3.5[/tex] represent the population standard deviation

[tex]n=49[/tex] sample size  

[tex]\mu_o =21[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the average age of the evening students is significantly different from 21, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 21[/tex]  

Alternative hypothesis:[tex]\mu \neq 21[/tex]  

The statistic is given by:

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

Calculate the statistic

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

[tex]z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4[/tex]    

P-value

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

[tex]p_v =2*P(z>4)=0.0000633[/tex]  

Conclusion  

When we compare the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can reject the null hypothesis at 10% of significance. So the  the true mean is difference from 21 at this significance level.

Final answer:

To determine whether the average age of the evening students is significantly different from 21, we can conduct a hypothesis test using a z-test with a known population standard deviation.

Explanation:

To determine whether the average age of the evening students is significantly different from 21, we can conduct a hypothesis test.

First, we need to state our hypotheses:

Null hypothesis (H0): The average age of the evening students is equal to 21.

Alternative hypothesis (Ha): The average age of the evening students is not equal to 21.

Since the population standard deviation is known, we can use a z-test. We calculate the test statistic using the formula:

z = (sample mean - hypothesized mean) / (population standard deviation / sqrt(sample size))

Once we have the test statistic, we can compare it to the critical value at a significance level of 0.1. If the test statistic falls within the critical region, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

In this case, the test statistic is 2.8571, which falls outside the critical region. Therefore, we reject the null hypothesis and conclude that the average age of the evening students is significantly different from 21.

A bag contains 23 coins, some dimes and some quarters. The total amount of money in the bag is $2.75. How many dimes and how many quarters are

in the bag?

X dimes

X quarters

Answers

Step-by-step explanation:

A bag contains 23 coins, some dimes and some quarters. The total amount of money in the bag is $2.75.

Let d be the  number of dimes

and q be the number of quarters

Total cons = 23

so equation becomes

[tex]d+q= 23\\[/tex]

[tex]q=23-d[/tex]

1 dime = 10 cents

and 1 quarter =25 cents

The total amount of money in the bag is $2.75

2.75= 275 cents

[tex]10d+25q=275[/tex]

solve the equation for d  and q

Plug in [tex]q=23-d[/tex] for q in second equation

[tex]10d+25q=275\\10d+25(23-d)=275\\10d+575-25d=275\\[/tex]

combine like terms and subtract 575 from both sides

[tex]10d+575-25d=275\\-15d=275-575\\-15d=-300\\[/tex]

divide both sides by -15

d=20

so the number of dimes = 20

[tex]q=23-d\\q=23-20\\q=3[/tex]

Number of quarter = 3

Answer:

20 dimes

3 quarters

Examine the results of a study1 investigating whether fast food consumption increases one’s concentration of phthalates, an ingredient in plastics which has been linked to multiple health problems including hormone disruption. The study included 8877 people who recorded all the food they ate over a 24-hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in ng/mL) in the urine: DEHP and DiNP. Find a confidence interval for the difference, , in mean concentration between people who have eaten fast food in the last 24 hours and those who haven’t. The mean concentration of DEHP in the 3095 participants who had eaten fast food was with while the mean for the 5782 participants who had not eaten fast food wasx¯N=59.1 with sN=152.1.

Answers

The 95% confidence interval for the difference in mean concentration between people who have eaten fast food and those who haven't is approximately  16.6  to  32.4  ng/mL.

To find the 95% confidence interval for the difference in mean concentration between people who have eaten fast food and those who haven't, we can use the formula for the confidence interval for the difference between two means:

[tex]\[ \text{CI} = (\bar{X}_1 - \bar{X}_2) \pm Z \times \sqrt{\frac{{S_1^2}}{{n_1}} + \frac{{S_2^2}}{{n_2}}} \][/tex]

Where:

[tex]\( \bar{X}_1 \) and \( \bar{X}_2 \)[/tex] are the sample means of the two groups.

[tex]\( S_1 \) and \( S_2 \)[/tex] are the sample standard deviations of the two groups.

[tex]\( n_1 \) and \( n_2 \)[/tex] are the sample sizes of the two groups.

Z is the critical value from the standard normal distribution for the desired confidence level.

Given:

[tex]\( \bar{X}_1 = 83.6 \)[/tex] (mean concentration of DEHP for people who have eaten fast food)

[tex]\( S_1 = 194.7 \)[/tex] (standard deviation of DEHP for people who have eaten fast food)

[tex]\( n_1 = 3095 \)[/tex] (sample size of people who have eaten fast food)

[tex]\( \bar{X}_2 = 59.1 \)[/tex] (mean concentration of DEHP for people who haven't eaten fast food)

[tex]\( S_2 = 152.1 \)[/tex] (standard deviation of DEHP for people who haven't eaten fast food)

[tex]\( n_2 = 5782 \)[/tex] (sample size of people who haven't eaten fast food)

First, we need to calculate the standard error of the difference in means:

[tex]\[ SE = \sqrt{\frac{{S_1^2}}{{n_1}} + \frac{{S_2^2}}{{n_2}}} \][/tex]

Then, we'll find the critical value Z for a 95% confidence interval, which corresponds to  Z = 1.96  for a two-tailed test.

Finally, we'll plug in the values to calculate the confidence interval.

Let's do the calculations:

First, let's calculate the standard error of the difference in means:

[tex]\[ SE = \sqrt{\frac{{194.7^2}}{{3095}} + \frac{{152.1^2}}{{5782}}} \]\[ SE \approx \sqrt{\frac{{37936.09}}{{3095}} + \frac{{23131.41}}{{5782}}} \]\[ SE \approx \sqrt{12.2604 + 4.0006} \]\[ SE \approx \sqrt{16.261} \]\[ SE \approx 4.032 \][/tex]

Now, we'll find the critical value  Z  for a 95% confidence interval, which corresponds to  Z = 1.96  for a two-tailed test.

Finally, we'll calculate the confidence interval:

[tex]\[ \text{CI} = (83.6 - 59.1) \pm 1.96 \times 4.032 \]\[ \text{CI} = 24.5 \pm 1.96 \times 4.032 \]\[ \text{CI} = 24.5 \pm 7.9072 \][/tex]

Now, let's find the bounds of the confidence interval:

Upper Bound:  24.5 + 7.9072 = 32.4072

Lower Bound:  24.5 - 7.9072 = 16.5928

Rounded to one decimal place, the 95% confidence interval for the difference in mean concentration between people who have eaten fast food and those who haven't is approximately  16.6  to  32.4  ng/mL.

Question :

Is Fast Food Messing With Your Hormones? Examine the results of a study investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics which has been linked to multiple health problems including hormone disruption. The study included 8877 people who recorded all the food they ate over a 24-hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in ng/mL) in the urine: DEHP and DiNP. Find a 95% confidence interval for the difference, HlF MN, in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DEHP in the 3095 participants who had eaten fast food was XF = 83.6 with SF = 194.7, while the mean for the 5782 participants who had not eaten fast food was TN = 59.1 with SN = 152.1. Round your answers to one decimal place: The 95% confidence interval is ______ to _________.

The 95% confidence interval is 16.6 ng/mL to 32.4 ng/mL.

To find the 95% confidence interval for the difference in mean concentration of DEHP between people who have eaten fast food in the last 24 hours and those who haven't, we use the formula for the confidence interval of the difference between two means with unequal variances (often referred to as Welch's t-test).

Given:

Mean concentration for fast food consumers [tex](\( \bar{X}_F \))[/tex] = 83.6 ng/mL

- Standard deviation for fast food consumers [tex](\( S_F \))[/tex] = 194.7 ng/mL

- Sample size for fast food consumers [tex](\( n_F \))[/tex] = 3095

- Mean concentration for non-fast food consumers [tex](\( \bar{X}_N \))[/tex] = 59.1 ng/mL

- Standard deviation for non-fast food consumers [tex](\( S_N \))[/tex] = 152.1 ng/mL

- Sample size for non-fast food consumers [tex](\( n_N \))[/tex] = 5782

The formula for the 95% confidence interval for the difference in means [tex](\( \mu_F - \mu_N \))[/tex] is:

[tex]\[ (\bar{X}_F - \bar{X}_N) \pm t^* \sqrt{\frac{S_F^2}{n_F} + \frac{S_N^2}{n_N}} \][/tex]

First, we calculate the standard error (SE):

[tex]\[ SE = \sqrt{\frac{S_F^2}{n_F} + \frac{S_N^2}{n_N}} \][/tex]

Plugging in the values:

[tex]\[ SE = \sqrt{\frac{194.7^2}{3095} + \frac{152.1^2}{5782}} \]\[ SE = \sqrt{\frac{37926.09}{3095} + \frac{23133.41}{5782}} \]\[ SE = \sqrt{12.253 + 4.001} \]\[ SE = \sqrt{16.254} \]\[ SE = 4.03 \][/tex]

Next, we find the degrees of freedom (df) using the formula for Welch's t-test:

[tex]\[ df \approx \frac{\left( \frac{S_F^2}{n_F} + \frac{S_N^2}{n_N} \right)^2}{\frac{\left( \frac{S_F^2}{n_F} \right)^2}{n_F-1} + \frac{\left( \frac{S_N^2}{n_N} \right)^2}{n_N-1}} \]\[ df \approx \frac{\left( 12.253 + 4.001 \right)^2}{\frac{12.253^2}{3094} + \frac{4.001^2}{5781}} \]\[ df \approx \frac{16.254^2}{\frac{150.13}{3094} + \frac{16.008}{5781}} \]\[ df \approx \frac{264.188}{0.0485 + 0.0028} \]\[ df \approx \frac{264.188}{0.0513} \]\[ df \approx 5149 \][/tex]

For a large [tex]\( df \), the \( t^* \)[/tex] value for a 95% confidence interval is approximately 1.96.

Now, we can calculate the confidence interval:

[tex]\[ (\bar{X}_F - \bar{X}_N) \pm t^* \cdot SE \]\[ (83.6 - 59.1) \pm 1.96 \cdot 4.03 \]\[ 24.5 \pm 7.9 \][/tex]

So, the 95% confidence interval for the difference in mean concentration of DEHP between people who have eaten fast food in the last 24 hours and those who haven't is:

[tex]\[ 16.6 \text{ to } 32.4 \][/tex]

The 95% confidence interval is 16.6 to 32.4 ng/mL.

Complete  question- Is Fast Food Messing With Your Hormones? Examine the results of a study investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics which has been linked to multiple health problems including hormone disruption. The study included 8877 people who recorded all the food they ate over a 24-hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in ng/mL) in the urine: DEHP and DiNP. Find a 95% confidence interval for the difference, HlF MN, in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DEHP in the 3095 participants who had eaten fast food was XF = 83.6 with SF = 194.7, while the mean for the 5782 participants who had not eaten fast food was TN = 59.1 with SN = 152.1. Round your answers to one decimal place: The 95% confidence interval is ______ to _________.

Jimmy’s Delicatessen sells large tins of Tom Tucker’s Toffee. The deli uses a periodic review system, checking inventory levels every 10 days, at which time an order is placed for more tins. Order lead time is 3 days. Average daily demand is 7 tins, so average demand during the reorder period and order lead time (13 days) is 91 tins. The standard deviation of demand during this same 13- day period is 17 tins. Calculate the restocking level. Assume the desired service level is 90% percent.

Answers

Answer:

The restocking level is 113 tins.

Step-by-step explanation:

Let the random variable X represents the restocking level.

The average demand during the reorder period and order lead time (13 days) is, μ = 91 tins.

The standard deviation of demand during this same 13- day period is, σ = 17 tins.

The service level that is desired is, 90%.

Compute the z-value for 90% desired service level as follows:

[tex]z_{\alpha}=z_{0.10}=1.282[/tex]

*Use a z-table for the value.

The expression representing the restocking level is:

[tex]X=\mu +z \sigma[/tex]

Compute the restocking level for a 90% desired service level as follows:

[tex]X=\mu +z \sigma[/tex]

   [tex]=91+(1.282\times 17)\\=91+21.794\\=112.794\\\approx 113[/tex]

Thus, the restocking level is 113 tins.

Final answer:

The restocking level for Jimmy's Delicatessen is calculated using the average demand, standard deviation, desired service level, and the z-score for a 90% service level, resulting in a restocking level of 113 tins.

Explanation:

Calculating the Restocking Level for Jimmy's Delicatessen

To calculate the restocking level, we need to use the information given about the average demand, the standard deviation of demand, and the desired service level. The average demand during the reorder period and order lead time (13 days) is 91 tins. Given the standard deviation of 17 tins and a 90% service level, we would typically look up the z-value that corresponds to a 90% service level in a standard normal distribution table, which is approximately 1.28.

Now, to find the restocking level, we use the formula: Restocking Level = Average Demand + (Z-score * Standard Deviation). Plugging in the numbers, we get:

Restocking Level = 91 tins + (1.28 * 17 tins) = 91 + 21.76 = 112.76 tins.

Therefore, the restocking level should be rounded up to 113 tins to ensure that there is a 90% probability that the stock on hand will be sufficient until the next delivery arrives.

slope of the line that passes through (3,14) and (10,6)

Answers

Answer:

-8/7

Step-by-step explanation:

2. In a random sample of 100 people, the correlation between amount of daily exercise and weight was found to be –.21. What would be the likely effect on the absolute value of the correlation coefficient under the following circumstances? (Hint: would r be greater or smaller? Why?) a. The sample is restricted to people who weighed less than 180 pounds

Answers

Answer:

Correlation coefficient 'r' would be lower.

Step-by-step explanation:

Correlation is co movement relationship between two variables.

Correlation coefficient 'r' is positive, when variables move in same direction. 'r' is negative when variables move in opposite direction. So, 'r' lies between -1 (perfect negative correlation) & +1 (perfect positive correlation). High 'r' magnitude reflects strong correlation between the variables, Low 'r' reflects  weak weak correlation between the variables.

Correlation studied between amount of daily exercise and weight  : It is negative as exercise & weight are negatively correlated - more exercise, less weight & less exercise, more weight. 'r' is given = -0.21

If sample is restricted to people weighing less than 180 pounds : It would lead to fall in 'r'. Such because these low weight people are likely to have good natural metabolic rate, naturally slim body physique / figure. So, in their case, exercise & body weight are likely to be less (weakly) correlated than normal case.

Solve.
A standard coffee mug has a capacity of 16 fluid ounces.
If Annie needs to fill 42 mugs with coffee, how many total quarts of coffee does she need?
Annie needs
quarts of coffee.​

Answers

Answer: 21 quarts. Annie will need a total of 21 quarts of coffee.

Step-by-step explanation: 16 fl oz = 0.5 quarts. 0.5 * 42 =21

21 quartz of coffee needed to 42 mugs.

what is Unitary Method?

The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.

For example, Let's say Ram spends 36 Rs. for a dozen (12) bananas.

12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.

As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.

This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.

Given:

Capacity of Coffee mug= 16 fluid ounce

1 fluid ounce = 0.03125 liquid quartz

16 fluid ounce = 16 x 0.03125

                       = 0.5 quartz

So, for 42 mugs the amount coffee needed

= 42 x 0.5

= 21 quartz

Hence, 21 quartz of coffee needed to 42 mugs.

Learn more about unitary method here:

https://brainly.com/question/22056199

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There were 5 1/3 jars of pickles. Ann and her friends ate 1 1/3 jars. How many jars of pickles are left? *

Answers

Answer:5/3 jars left

Step-by-step explanation:

Convert the total number of jars to improper fraction from mixed fraction

The subtract the total number of jars from the total number of empty jars

16/3 -11/3

5/3

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