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3 years ago

What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as

Mathematics

1 answer:

Andrej [43]3 years ago

5 0

Complete question :

Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.

What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.

Answer:

0.56 ; 0.60

Step-by-step explanation:

From The attached Venn diagram :

C = attend college ; J = has a job

P(C) = (35+45)/300 = 80/300 = 8/30

P(J) = (30+45)/300 = 75/300 = 0.25

P(C n J) = 45 /300 = 0.15

1.)

P(J | C) = P(C n J) / P(C)

P(J | C) = 0.15 / (8/30)

P(J | C) = 0.5625 = 0.56

2.)

P(C | J) = P(C n J) / P(J)

P(C | J) = 0.15 / (0.25)

P(C | J) = 0.6 = 0.60

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1. Construct a table of values of the following functions using the interval of 5

Morgarella [4.7K]

Complete Question:

Construct a table of values of the following functions using the interval of -5 to 5.

$g(x) = \frac{x^3 + 3x - 5}{x^2}$

Answer:

See Explanation

Step-by-step explanation:

Required

Construct a table with the given interval

When $x = -5$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-5) = \frac{-5^3 + 3(-5) - 5}{-5^2}$

$g(-5) = \frac{-125 -15 - 5}{25}$

$g(-5) = \frac{-145}{25}$

$g(-5) = -5.8$

When $x = -4$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-4) = \frac{-4^3 + 3(-4) - 5}{-4^2}$

$g(-4) = \frac{-64 -12 - 5}{16}$

$g(-4) = \frac{-81}{16}$

$g(-4) = -5.0625$

When $x = -3$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-3) = \frac{-3^3 + 3(-3) - 5}{-3^2}$

$g(-3) = \frac{-27 -9 - 5}{9}$

$g(-3) = \frac{-41}{9}$

$g(-3) = -4.56$

When $x = -2$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-2) = \frac{-2^3 + 3(-2) - 5}{-2^2}$

$g(-2) = \frac{-8 -6 - 5}{4}$

$g(-2) = \frac{-19}{4}$

$g(-2) = -4.75$

When $x = -1$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-1) = \frac{-1^3 + 3(-1) - 5}{-1^2}$

$g(-1) = \frac{-1 + 3 - 5}{1}$

$g(-1) = \frac{-3}{1}$

$g(-1) = -3$

When $x = 0$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(0) = \frac{0^3 + 3(0) - 5}{0^2}$

$g(0) = \frac{0 + 0 - 5}{0}$

$g(0) = \frac{- 5}{0}$

g(0) = undefined

When $x = 1$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(1) = \frac{1^3 + 3(1) - 5}{1^2}$

$g(1) = \frac{1 + 3 - 5}{1}$

$g(1) = \frac{-1}{1}$

$g(1) = 1$

When $x = 2$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(2) = \frac{2^3 + 3(2) - 5}{2^2}$

$g(2) = \frac{8 + 6 - 5}{4}$

$g(2) = \frac{9}{4}$

$g(2) = 2.25$

When $x = 3$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(3) = \frac{3^3 + 3(3) - 5}{3^2}$

$g(3) = \frac{27 + 9 - 5}{9}$

$g(3) = \frac{31}{9}$

$g(3) = 3.44$

When $x = 4$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(4) = \frac{4^3 + 3(4) - 5}{4^2}$

$g(4) = \frac{64 + 12 - 5}{16}$

$g(4) = \frac{71}{16}$

$g(4) = 4.4375$

When $x = 5$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(5) = \frac{5^3 + 3(5) - 5}{5^2}$

$g(5) = \frac{125 + 15 - 5}{25}$

$g(5) = \frac{135}{25}$

$g(5) = 5.4$

Hence, the complete table is:

x ---- g(x)

-5 --- -5.8

-4 --- -5.0625

-3 --- -4.56

-2 --- -4.75

-1 --- -3

0 -- Undefined

1 --- 1

2 -- 2.25

3 --- 3.44

4 --- 4.4375

5 --- 5.4

7 0

3 years ago

The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean o

Fynjy0 [20]

Answer:

a) 5.6 years

b) 0.4478

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 5.6 years

Standard Deviation, σ = 1.9 years

Sample size, n = 18

We are given that the distribution of amounts of time is a bell shaped distribution that is a normal distribution.

a) Mean of each sample.

The best estimator of the sample mean is the population mean itself. Thus, we have:

$\mu_{x} = \mu = 5.6$

Thus, the sample mean is 5.6 years.

b) Standard error

Formula:

$S.E = \dfrac{\sigma}{\sqrt{n}}$

Putting values, we have.

$S.E = \dfrac{1.9}{\sqrt{18}} = 0.4478$

Thus, the standard error is 0.4478.

7 0

3 years ago

Which number is three million times larger than 2.3 x 104 ? A) 2.3 x 107 B) 2.3 x 108 C) 6.9 x 109 D) 6.9 x 1010

Strike441 [17]

Your answer is D 6.9*1010 because... Q: 2.3*104=239.2 A 2.3*107=246.1 B 2.3*108=248.4 C 6.9*109=752.1 D 6.9*1010=6,969 >>>> Your answer Your answer is D.

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zheka24 [161]

Answer:

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Step-by-step explanation:

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ELEN [110]

Answer:

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