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2 years ago

Which of this is equivilent to the expression s+s+s+s+12

Mathematics

1 answer:

Tanya [424]2 years ago

6 0

4s + 12 or 4(s + 3) would be equivalent to the expression!

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How I can answer this question, NO LINKS, If you answer correctly I will give u brainliest!

Vikentia [17]

Answer:

$\huge\boxed{\sf n = 7}$

Step-by-step explanation:

<u>Given equation is:</u>

$(7^2)^4=n^8$

Multiply 2 to 4

$7^8=n^8$

While comparing both sides, we get:

7 = n

n = 7

$\rule[225]{225}{2}$

Hope this helped!

7 0

2 years ago

Read 2 more answers

Can you help me please help me

pashok25 [27]

Answer:

I can`t really see the question that well, but if I understand correctly here`s the first few.

1. Subtraction

2. Addition

3. Addition

4. Subtraction

5. Addition

6. Subtraction

It looks like its asking what is happening in those situations (is the number being added or subtracted). Sorry if that`s not much help

4 0

3 years ago

Please answer the side questions and the "#justkeepswimming" thank you if you do!!

fenix001 [56]

#just keep swimming I don’t know what you mean

4 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

8 0

3 years ago

Identify the equation that represents this solution & the correct solution.

Hitman42 [59]

The answer is A. d-17= 22; d=39

6 0

3 years ago