Ask question

Ask question

All categories

2 years ago

15

can someone help me im tryna finish as. much as pre-algebra as i can im leaving to vacation and i need to finsish allot of math!

someone help :C

1 answer:

konstantin123 [22]2 years ago

7 0

Answer:

i only know the answer for the first one, im sorry, its 30

Step-by-step explanation:

a=h b/2 if you need help with finding the area of something again use this, Geometry calculator

You might be interested in

Round the number of connected computers 42,199,279 to the nearest ten million

Flura [38]

10, 000, 000 = ten million
40, 000 ,000 is the answer for the nearest ten million

3 0

3 years ago

Read 2 more answers

Can someone pls help ASAP :) I’ll give brainiest

Anna71 [15]

Answer:

D. (2x)(3x) + 5(3x - 2)

<em>good luck, i hope this helps :) </em>

<em />

5 0

3 years ago

Anyone know the answer?

sattari [20]

The answer is c because k and m are the same so l and n are also the same

7 0

3 years ago

The price-earnings ratio for firms in a given industry follows the normal distribution. In this industry, a firm whose price-ear

lesantik [10]

Answer:

The correct answer is

A. 16%

Step-by-step explanation:

The zscore is used to solve problems of normally distributed samples.

This score indicates the percentage of a certain measure. It can be found looking at the z table.

For example, if a measure has a percentile of 75%, it is in the 100%-75% = highest 25 percent of all the measures.

$Z = 1$ has a pvalue 0.8413. So it is in the highest 1-0.8413 = 0.1587 = 16% percent of firms in the industry.

The correct answer is

A. 16%

4 0

3 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

5 0

3 years ago