Forty percent of all Americans who travel by car look for gas stations and food outlets that are close to or visible from the highway. Suppose a random sample of n=25 Americans who travel by car are asked how they determine where to stop for food and gas. Let x be the number in the sample who respond that they look for gas stations and food outlets that are close to or visible from the highway. a. What are the mean and variance of x?
b. Calculate the interval μ±2σμ±2σ. What values of the binomial random variable x fall into this interval?
c. Find P(6≤≤x$\leq$14). How does this compare with the fraction in the interval μ±2σμ±2σ for any distribution? For mound-shaped distributions?

Answer:

The correct solutions are (6, -2).

Step-by-step explanation:

For the first equation, rearrange to make x the subject.

3x + 4y = 10

Divide the whole equation by 3 to isolate x:

Now substitute this into the second equation:

Subtract 20 from both sides:

20 - 10y = 40

20 - 10y - 20 = 40 - 20

-10y = 20

Divide both sides by 2:

-10y ÷ 10 = 20 ÷ 10

-y = 2 ∴ y = -2

Plug this value back into the first equation:

3x + 4y = 10

3x + 4(-2) = 10

3x + (-8) = 10

3x - 8 = 10

Add 8 to both sides:

3x - 8 + 8 = 10 + 8

3x = 18

Divide both sides by 3:

3x ÷ 3 = 18 ÷ 3

x = 6

Therefore, the correct solutions are (6, -2).

Hope this helps!

The sum of A and B can have no carry. If we assume A ≠ B ≠ C, then you can have ...

... (A, B, C) ∈ {(1, 2, 3), (2, 1, 3), (1, 3, 4), (3, 1, 4), (1, 4, 5), (4, 1, 5), (1, 5, 6), (5, 1, 6), (1, 6, 7), (6, 1, 7), (1, 7, 8), (7, 1, 8), (1, 8, 9), (8, 1, 9), (2, 3, 5), (3, 2, 5), (2, 4, 6), (4, 2, 6), (2, 5, 7), (5, 2, 7), (2, 6, 8), (6, 2, 8), (2, 7, 9), (7, 2, 9), (3, 4, 7), (4, 3, 7), (3, 5, 8), (5, 3, 8), (3, 6, 9), (6, 3, 9), (4, 5, 9), (5, 4, 9)}

Answer:

{ 1,5}

Step-by-step explanation:

The intersection is what the two sets have in common

{1, 5, 10, 15}∩ {1, 3, 5, 7}

= { 1,5}

Answer:

{1,5}

Step-by-step explanation:

The intersection of the sets are all of the numbers that appear in both sets. In this case, the only numbers that appear in both are 1 and 5.

Answer:  x = 21

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Explanation:

The exponent rules are

• a^b * a^c = a^(b+c) .... lets call this rule 1
• (a^b)/(a^c) = a^(b-c) .... lets call this rule 2
• (a^b)^c = a^(b*c) ......... lets call this rule 3

The expression on the right hand side simplifies to c^(-1). We add the exponents to get this. So we're using rule 1 mentioned above.

We'll keep this in mind for later.

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On the left hand side (c^5)^4 becomes c^(20) because we multiply the exponents (rule 3)

Then (c^20)/(c^x) becomes c^(20-x). We subtract exponents here (rule 2).

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After using those exponent rules, the original equation turns into

c^(20-x) = c^(-1)

The bases are both c, so the exponents must be equal

20-x = -1

-x = -1-20

-x = -21

x = 21

The Z-test of a population mean is used here because it is given that the population standard deviation is $75. The correct option is a).

Given :

• The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75.
• A sample of 30 days reveals a daily average revenue of $625.

Let is the daily average revenue. So:

Null Hypothesis is,

The Alternate Hypothesis is,

Here, the Z-test of a population mean is used because it is given that the population standard deviation is $75.

where is the sampleaverage revenue, is the standard deviation and n is the sample size.

Therefore, the correct option is a).

For more information, refer to the link given below:

brainly.com/question/13245779

Answer:

We would use Z-test of a population mean.

Step-by-step explanation:

We are given that an entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625.

And we have to test the hypothesis that the daily average revenue was $675.

Firstly, as we know that testing is always done on the population parameter.

So, let = daily average revenue over the pat 5 years

So, Null Hypothesis, : = $675

Alternate Hypothesis, : $675

Here, null hypothesis states that the owner's claim of average daily revenue was $675 over the past 5 years is correct.

And alternate hypothesis states that the owner's claim of average daily revenue was $675 over the past 5 years is not correct.

The test statistics that will be used here is Z-test of a population mean because here we have knowledge of population standard deviation of $75.

   Test statistics =  ~ N(0,1)

where, = sample average revenue = $625

            = population standard deviation = $75

             n = sample of days = 30

Therefore, to test the null hypothesis that the daily average revenue was $675, we should use Z-test of a population mean.

Answer:

1. A. You hire the applicant when the applicant will not be an asset to the company.

2. C. You do not hire the applicant when the applicant will be an asset to the company.

Step-by-step explanation:

1. The type I error happens when the null hypothesis is rejected when it is true, in this way we know that the null hypothesis is that the new employee will not be active for the company, so option B is rejected, because it refers that the Applicant if he will be active or for the company, option C is rejected because the inactive employee is rejected, accepting the null hypothesis, option D is rejected because the contracted applicant if active, so the correct answer is A, in which the inactive applicant is hired.

2.

we know that the type II error occurs when the null hypothesis is accepted, being this false, we know that the null hypothesis is to hire an inactive applicant for the company, so option A is not correct, in which the null hypothesis is accepted taking it as true, option B is rejected, in which the contract is made to an active applicant, so the null hypothesis is false and option D is rejected, in which the null hypothesis is rejected, therefore the correct answer It is the C in which the active applicant is not hired.

Answer:

1. Option A

2. Option C

Step-by-step explanation:

The null hypothesis is that the applicant will not be an asset to the company, thus you do not hire such applicant

The alternative hypothesis is that the applicant will be an asset to the company and you then hire such applicant.

A type I error occurs when the researcher rejects the null hypothesis when true.

A type II error occurs when the researcher fails to reject the null hypothesis when it is not true.

1. Type I error:

You hire the applicant when the applicant will not be an asset to the company

2. Type II error:

You do not hire the applicant when the applicant will be an asset to the company.

3. Type I error because you rejected the null hypothesis to not hire when the applicant will not be an asset to the company.