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3 years ago

8

A manger is scudeuleing interviews for job applicants. She has 8 time slots and 6 applicants to interview. How many diffrent int

erview schedules are possible?

1 answer:

borishaifa [10]3 years ago

8 0

Answer:

There are 2 different schedules

Step-by-step explanation:

8 time slots and 6 interviews 4 are the same meaning 2 are different

You might be interested in

The diagram below shows a student using a clinometer to measure the height of a flagpole. When the student stands 40 feet away f

AVprozaik [17]

The correct answer is the flagpole is <span>33 feet high</span>.

Explanation:
Please refer to the attached picture.

We know:
CD = 40 feet
AC = 5 feet
∠BDC = α = 35°

Using trigonometry, we know that the definition of the tangent of an angle is the ratio between the opposite side and the adjacent side, therefore:
tan α = BC / CD

Solving for BC:
BC = CD · <span>tan α
= 40 </span>· tan (35)
= 28 feet

In order to find the height of the flagpole, we need to add the distance of the clinometer from the ground:
AB = BC + AC
= 28 + 5
= 33

Hence, the flagpole is 33 feet high.

5 0

3 years ago

WHICH LINE HAS A SLOPE OF 0 AND GOES THROUGH (5, -8)

arlik [135]

The slope<span> of a horizontal </span>line<span>. A horizontal </span><span>line has slope 0

</span>

3 0

3 years ago

. Noel buys 12 balloons and 35 paper cups for his birthday party. What is the ratio of balloons to paper cups? 3.

Varvara68 [4.7K]

Answer:

12:35

Step-by-step explanation:

7 0

2 years ago

Twelve percent of U.S. residents are in their forties. Consider a group of nine U.S. residents selected at random. Find the prob

miskamm [114]

Answer:

The probability is 28 % ( approx )

Step-by-step explanation:

Since, there are only two possible outcomes in every trails ( residents are in their forties or not ),

So, this is a binomial distribution,

Binomial distribution formula,

Probability of success in x trials,

$P(x)=^nC_x p^x q^{n-x}$

Where, $^nC_x=\frac{n!}{x!(n-x)!}$

p and q are probability of success and failure respectively and n is the total number of trials.

Let X be the event of resident who are in his or her forties.

Here, p = 12% = 0.12,

⇒ q = 1 - p = 1 - 0.12 = 0.88,

n = 9,

Hence, the probability that two or three of the people in the group are in their forties = P(X=2) + P(X=3)

$^9C_2 (0.12)^2 (0.88)^{9-2}+^9C_3 (0.12)^3 (0.88)^{9-3}$

$=36(0.12)^2 (0.88)^7+84(0.12)^3 (0.88)^6$

$=0.279266611163$

$\approx 0.28$

$=28\%$

7 0

3 years ago

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma

nata0808 [166]

Answer:

a) $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c) $P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d) $P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e) $P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

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".

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

$X \sim N(110000,40000)$

Where $\mu=110000$ and $\sigma=40000$

If we select a sample size of n =35 the standard error is given by:

$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

$P(X > 112000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

$P(X > 100000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

$P(100000$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

8 0

3 years ago