Which expression is the factored form of -4.5n+3?

Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mea

QveST [7]

A.
$\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)$

Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.

$\mathbb P(Z>-2)=\mathbb P(|Z|2)=0.95+0.025=0.975$

b.
$\mathbb P(1100$

About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.

c. You're looking for $k$ such that

$\mathbb P(X>k)=0.10$

Since

$\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10$

occurs for $k^*\approx1.2816$ , it follows that

$\dfrac{k-1200}{100}=1.2816\implies k\approx1328$

So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.

3 0

4 years ago

Find three consecutive odd integers whose sum is 51

Sedbober [7]

Answer: 15, 17 and 19

Step-by-step explanation:

15 + 17 + 19 = 51

3 0

3 years ago

The test score for a class are shown. What is the average test score? 79,80,92,92,81,100,88,98,71,100,91,90

Oksanka [162]

Answer:

The average score is equal to 88.5

Step-by-step explanation:

we know that

An average test score is the sum of all the scores on an assessment divided by the number of test-takers.

so

The sum is equal to

$Sum=[79+80+92+92+81+100+88+98+71+100+91+90]=1,062$

The number of test-takers is n=12

The average score is equal to

$\frac{S}{n}$

substitute

$\frac{1,062}{12}=88.5$

5 0

3 years ago

Suppose the ice Q scores in one region are normally distributed within a standard deviation of 13. Suppose also that exactly 51%

elena-14-01-66 [18.8K]

Answer:

$z=-0.025$

And if we solve for $\mu$ we got

$\mu=100 +0.025*13=100.325$

Step-by-step explanation:

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".

Solution to the problem

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

$X \sim N(\mu,13)$

Where $\mu=?$ and $\sigma=13$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

For this case we know these conditions:

$P(X>100)=0.51$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.49 of the area on the left and 0.51 of the area on the right it's z=-0.025. On this case P(Z<-0.025)=0.49 and P(z>-0.025)=0.51

If we use condition (b) from previous we have this:

$P(X$