Steven got a one time $90 bonus. He got paid $330. How much did he make before his bonus?

One type of insect is 0.0052 meters long. what is this length in scientific notation?

pochemuha

Your answer would be:
5.2*10^(-3)

hope this helps!
happy thanksgiving

4 0

3 years ago

Read 2 more answers

8. 36.8 = [11.5 – (2.5 X 3)]2

Anestetic [448]

Answer:

$[11.5-(2.5*3)]^2=16$

Step-by-step explanation:

We want to evaluate $[11.5-(2.5*3)]^2$

Let us evaluate within the parenthesis first:

$[11.5-(2.5*3)]^2=[11.5-(7.5)]^2$

$\implies [11.5-(2.5*3)]^2=[11.5-7.5]^2$

We again subtract within the bracket to obtain:

$[11.5-(2.5*3)]^2=[4]^2$

This finally gives us:

$[11.5-(2.5*3)]^2=16$

6 0

3 years ago

Find the indicated probability or percentage for the sampling error. The distribution of weekly salaries at a large company is r

Flauer [41]

Answer:

The probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

Step-by-step explanation:

According to the Central Limit Theorem if we have a non-normal population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the distribution of sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The information provided is:

$\mu=\$1000\\\sigma=\$370\\n=80$

As n = 80 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean weekly salaries.

Let $\bar X$ represent the sample mean weekly salaries.

The distribution of $\bar X$ is: $\bar X\sim N(\$1000,\ \$41.37)$

Now we need to compute the probability of the sampling error made in estimating the mean weekly salary to be at most $75.

The sampling error is the the difference between the estimated value of the parameter and the actual value of the parameter, i.e. in this case the sampling error is, $|\bar X-\mu|= 75$ .

Compute the probability as follows:

$P(-75$

$=P(-1.81$

Thus, the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

3 0

3 years ago

Arrange the steps in the correct order to express cos3x in terms of cosx cos(2x)cos(x)-sin(2x)(x) [1-sin^2(x)]cos(x)-[2sin(x)cos

noname [10]

cos(3x) = cos(2x+x)

= cos(2x) cos (x) -sin(2x)sin(x)

$=(1-2sin^2(x)) cos(x)-(2 sin(x) cos(x)) sin(x)$