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Help me almost done!

Andreyy89

Cos 60° is ½

and sin 60° is √3/2

so it is true....

that table may help you

6 0

3 years ago

Graph shows the solution to the following inequality 40 - 6 is less than 18

Veseljchak [2.6K]

Step 1) Add 6 to both sides
Step 2) Divide both sides by 4

Following those steps above, you'll have this
4d - 6 < 18
4d - 6 + 6 < 18 + 6 .... applying step 1
4d + 0 < 24
4d < 24
4d/4 < 24/4 .... applying step 2
d < 6

So d can be any number that is less than 6. To graph this, we draw a number line, plot an open circle at 6 and shade to the left of the open circle. The open circle indicates to exclude the endpoint. The graph I'm describing is choice D

Answer: Choice D

3 0

3 years ago

Read 2 more answers

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

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