PLEASE HELP

What is the answer to this problem??please factor completely 9r^3-64r^2+r-8

slamgirl [31]

Answer:

Wait do you mean 8r^3-64r^2+r-8?

Step-by-step explanation:

4 0

3 years ago

AB ≅CD, FA ≅ED

Rama09 [41]

This answer is the SAS therom because you have 2 congruent sides and then you have the perpendicular bisector, which is a right angle

7 0

3 years ago

Find the area of the composite figure

Kamila [148]

Take the two shapes separately:

Rectangle:
l x w = A
22 x 17 =?

Triangle:
.5b x h = A
.5(12) x 19 = ?

A of Triangle + A of Rectangle = total Area

6 0

4 years ago

Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed w

jok3333 [9.3K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The sales level that is the cut-off between quarters that are considers as "failure" and those that are not is evaluated as

$y = 6.3 \ million \ dollars$

Step-by-step explanation:

From the question we are told that the

Mean is $\mu = 8 \ million \ dollars$

Standard deviation is $\sigma = 1.3 \ million \ dollars$

Let Y be the random variable that denotes the sales made quarterly among health care information system

The normal distribution for this data is mathematically represented as

$Y$ ~ N $(\mu = 8 , \sigma^2 = 1.69)$

From the question we are told that a quarter is consider a failure by the company if the sales level that quarter in the bottom 10% of the of all quarterly sales

So

The the Probability of obtaining quarterly sales level that is equal to 10% of all quarterly sale (It is not below 10%) is mathematically represented as

$P[Y < y ] = 0.10$

This can be represented as a normal distribution in this manner

$P [ \frac{Y- \mu}{\sigma } < \frac{y -\mu}{\sigma} ] = 0.10$

Where $\frac{Y - \mu }{\sigma } = Z$ ~ $N(0 , 1)$

{This mean that this equal to the normal distribution between 0, 1 which is the generally range of every probability }

Therefor we have

$P[Z < \frac{y- 8}{1.3} ]$

The cumulative distribution function for a normal distribution of y is mathematically represented as

$\phi [\frac{y- 8}{1.3} ] = 0.10$

This is because a cumulative distribution function of a random value Y or a distribution of Y evaluated at y is the probability that Y will take will take a value that is less or equal to y

$\frac{y- 8}{1.3} = \phi ^{-1}(0.10)$

Calculating the inverse of the cumulative distribution function value of 0.10 which is negative the the critical value(Critical values determine what probability a particular variable will have when a sampling distribution is normal or close to normal.) of 0.01

$\phi ^{-1}(0.10) =- [1 - \frac{0.10}{2}]$