2. Evaluate the expression by substituting the given value(s):

Find the y-intercept of the line: -4x - 2y = 16

Georgia [21]

The y-intercept is (0, -8)
if you set the equation to slope intercept format you get

y=-2x-8

3 0

3 years ago

2+Inx=5 How to solve?

irga5000 [103]

Ln x=5-2
ln x=3
ln x is base e log x
so,
x=e^(3)
x=20.0855369232

4 0

3 years ago

Read 2 more answers

In a recent study on world happiness, participants were asked to evaluate their current lives on a scale from 0 to 10, where 0

uysha [10]

Answer:

a) A response of 8.9 represents the 92nd percentile.

b) A response of 6.6 represents the 62nd percentile.

c) A response of 4.4 represents the first quartile.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5.9

Standard Deviation, σ = 2.2

We assume that the distribution of response is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) We have to find the value of x such that the probability is 0.92

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.92$

Calculation the value from standard normal z table, we have,

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = 1.405\\x = 8.991 \approx 8.9$

A response of 8.9 represents the 92nd percentile.

b) We have to find the value of x such that the probability is 0.62

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.62$

Calculation the value from standard normal z table, we have,

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = 0.305\\x = 6.571 \approx 6.6$

A response of 6.6 represents the 62nd percentile.

c) We have to find the value of x such that the probability is 0.25

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.25$

Calculation the value from standard normal z table, we have,

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = -0.674\\x = 4.4172 \approx 4.4$

A response of 4.4 represents the first quartile.

4 0

3 years ago

Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,

Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that $p = 0.15$

Sample of 6:

This means that $n = 6$

What is the probability that there will be 0 or 1 defects in a sample of 6?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771$

$P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764$

77.64% probability that there will be 0 or 1 defects in a sample of 6.

5 0

3 years ago

I need help with number 15,16,and 17

777dan777 [17]

15. A=7, 16. V=9, 17. D=26

7 0

3 years ago