1. What kind of correlation is shown? (see #1 above)

The life of light bulbs is distributed normally. The standard deviation of the lifetime is 15 hours and the mean lifetime of a b

Anastaziya [24]

Answer:

0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 590 hours.

This means that $\sigma = 15, \mu = 590$

Find the probability of a bulb lasting for at most 605 hours.

This is the pvalue of Z when X = 605. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{605 - 590}{15}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.

7 0

3 years ago

the line -6+2y=10 is written in standard form. can you tell what the growth of the line is?Its y-intercept?

slamgirl [31]

Answer:

Growth: 3

y-intercept: 5

Step-by-step explanation:

1. Find the equation in slope-intercept form:

-6x+2y=10

2y=6x+10

y=3x+5

2. Then we can easily identify the growth and y-intercept.

Growth: 3

y-intercept: 5

6 0

3 years ago

Read 2 more answers

Which function represents the graph of h(x)=2|x+3|−1 after it is translated 2 units right?

harina [27]

$\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}$

$\bf ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

now with that template in mind

$\bf h(x)=\stackrel{A}{2}|\stackrel{B}{1}x+\stackrel{C}{3}|\stackrel{D}{-1}\qquad \qquad \stackrel{\textit{C=C-2 a translation to the right}}{h(x)=2|x\boxed{+3-2}|-1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill h(x)=2|x+1|-1~\hfill$