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3 years ago

34) Which equation is equivalent to: 3r=78+14 ?

Mathematics

1 answer:

Julli [10]3 years ago

8 0

It’s B. (Note: The image below isn’t mine but I can assure you it is B)

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Solve for f. show your work! 12=f-13-2

tankabanditka [31]

12=f-13-2
12=f-15
f=27

3 0

4 years ago

Read 2 more answers

Simplify the expression: w2 + 7w2 + 8w - 5 + 9 - 2w2

ololo11 [35]

<h3> $6w^2 +8w +4$ is the simplified expression</h3>

<em><u>Solution:</u></em>

Given that,

We have to simplify

$w^2 + 7w^2 + 8w -5+9-2w^2$

We can simplify the above expression by combining the like terms

Like terms are terms that has same variable with same exponent and same or different coefficient

From given,

$w^2 + 7w^2 + 8w -5+9-2w^2$

Group the like terms

$w^2 + 7w^2 -2w^2 + 8w - 5+9\\\\Combine\ the\ like\ terms\\\\6w^2 +8w -5+9\\\\Combine\ the\ constants\\\\6w^2 +8w +4$

Thus the given expression is simplified

4 0

3 years ago

4/5 y = 8 , how do we solve pls help fast its urgent, I will mark u brainliest

Butoxors [25]

Answer:

y = 10

Step-by-step explanation:

4/5 y = 8

4/5 y * 5/4 = 8 * 5/4 multiply by reciprocal

y = 10

4 0

4 years ago

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Daniel wants to do 312 practice problems to prepare for an exam. if he wants to do approximately the same number of problems eac

alexandr1967 [171]

Answer:

Step-by-step explanation:

6 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