Ask question

Ask question

All categories

3 years ago

7

Out of 431 applicants for a job, 142 have over 5 years of experience and 103 have over 5 years of experience and have a graduate

degree. Consider that 144 of the applicants have graduate degrees. What is the probability that a randomly chosen applicant has over 5 years of experience, given that the applicant has a graduate degree? Enter a fraction or round your answer to 4 decimal places, if necessary.

1 answer:

Savatey [412]3 years ago

3 0

Answer:

0.2390

Step-by-step explanation:

<u>Given</u>

Total applicants = 431
Applicants with graduate degree and over 5 years of experience = 103

Probability = favorable outcomes/total outcomes

<u>Required probability </u>

103/431 = 0.2390 rounded to 4 decimal places

You might be interested in

Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?

worty [1.4K]

Answer:

4

Step-by-step explanation:

$(x+y+z)(xy+xz+yz)=18$ Equation 1

$x^2(y+z)+y^2(x+z)+z^2(x+y)=6$ Equation 2

What is the value of $xyz$ where each variable represents a real number?

Let's expand equation 1:

$(x+y+z)(xy+xz+yz)=18$

$x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18$

Simplify each term if can:

$x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18$

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have $x^2$ and if I factored $x^2$ from first two terms I would have $x^2(y+z)$ which is the first term of left hand side of equation 2.

So let's see what happens if we gather the terms together that have the same variable squared together.

$x^2y+x^2z+y^2x+y^2z+z^2y+z^2x+xyz+xyz+xyz=18$

Factor the variable squared terms out of each binomial pairing:

$x^2(y+z)+y^2(x+z)+z^2(y+x)+xyz+xyz+xyz=18$

Replace the sum of those first three terms with what it equals which is 6 from the equation 2:

$6+xyz+xyz+xyz=18$

Combine like terms:

$6+3xyz=18$

Subtract 6 on both sides:

$3xyz=12$

Divide both sides by 3:

$xyz=4$

4 0

3 years ago

A random sample of 8 recent college graduates found that starting salaries for architects in New York City had a mean of $42,653

ycow [4]

Answer:

C. (36337.32, 48968.68)

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{9114}{\sqrt{8}} = 6315.68$

The lower end of the interval is the sample mean subtracted by M. So it is 42653 - 6315.68 = 36337.32.

The upper end of the interval is the sample mean added to M. So it is 42653 + 6315.68 = 48968.68.

So the correct answer is:

C. (36337.32, 48968.68)

3 0

3 years ago

???????? Why ir why not

denpristay [2]

yes because it passes the vertical line test

4 0

3 years ago

Read 2 more answers

Using number line what is 4-5

Vinil7 [7]

Answer:

<h2>-1</h2>

Step-by-step explanation:

4 - 5 = -1

Count:

4, 3, 2, 1, 0, <u>-1</u>

6 0

3 years ago

What is the <br> Number that the are talking about

anzhelika [568]

i can't comment so someone else is going to have to answer

but they need the whole page to answer this question.

4 0

3 years ago