Answer me pls I will mark brainliest worth 50 points. (sorry when I put 100 it goes to 50).

2 answers:

4 0

The typical salary at this company is $24,000, the median. Yes, it is misleading; the average is thrown off by the outlier value of $79,000.

il63 [147K]3 years ago

3 0

Answer:

Yes, because 6/8 employees recieve salaries of lower than 31500, and one of them get more than twice the salary than the average salary

You might be interested in

How do you write 0.334 in expanded form?

NikAS [45]

0.300+0.030+0.004 hope this helps

4 0

3 years ago

Find the limit of the function algebraically. (2 points) limit as x approaches zero of quantity x squared plus two x divided by

DIA [1.3K]

Answer:

DNE

Step-by-step explanation:

$\lim_{x \to \0} \frac{x^{2} +2x}{x^{4} }$

As you can see in the picture I attached to this, that as the limit goes to zero from the negative side, it approaches -∞ and from the positive side, it approaches ∞ . hence, the limit doesn't exist.

To show this algebraically, you have to imagine how a number divided by zero looks like, Just know the graph of 1/x and see how the limit to zero doesnot exist. I hope this helps!

7 0

3 years ago

Laser scanning for fish volume estimation. engineers design tanks for rearing commercial fish to minimize both the use of natura

AVprozaik [17]

Answer:

The 99% confidence interval would be given by (196.2;283.8)

We are 99% condident that the true mean is between 196.2 and 283.8

We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=240$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=15 represent the sample standard deviation

n=4 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=4-1=3$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,3)".And we see that $t_{\alpha/2}=5.84$