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defon
3 years ago
7

PLEASE ANSWER QUICKLY WILL MARK BRAINLIEST!!!!!!!!!!!!!!!!! Simplify

la1" title="4^{2}" alt="4^{2}" align="absmiddle" class="latex-formula">·4^{8}
Mathematics
2 answers:
pashok25 [27]3 years ago
8 0

Answer:

4^10/ 4 to the 10th power

Step-by-step explanation:

4 x 4 = 16

4x4x4x4x4x4x4x4= 65536

65536(16)= 1,048,576.

Simplified, it's 4^10 or 4 to the 10th power.

Hope this helps!

Elodia [21]3 years ago
6 0

Hey there! I'm happy to help!

When multiplying numbers with exponents, you add the numbers in the exponent. So, 4^2+4^8=4^1^0. If you plug this into your calculator, you get 1048576, which is 4^1^0.

Have a wonderful day! :D

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\sigma = 11.28 --- Standard deviation

Step-by-step explanation:

Given

See attachment for graph

Solving (a): Explain how the standard deviation is calculated.

<u>Start by calculating the mean</u>

To do this, we divide the sum of the products of grade and number of students by the total number of students;

i.e.

\bar x = \frac{\sum fx}{\sum f}

So, we have:

\bar x = \frac{50 * 1 + 57 * 2 + 60 * 4 + 65 * 3 +72 *3 + 75 * 12 + 77 * 10 + 81 * 6 + 83 * 6 + 88 * 9 + 90 * 12 + 92 * 12 +95 * 2 + 99 * 4 + 100 * 5}{1 + 2 + 4 + 3 +3 + 12 + 10 + 6 + 6 + 9 + 12 + 12 +2 + 4 + 5}

\bar x = \frac{7531}{91}

\bar x = 82.76

Next, calculate the variance using the following formula:

\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f}

i.e subtract the mean from each dataset; take the squares; add up the squares; then divide the sum by the number of dataset

So, we have:

\sigma^2 = \frac{1 * (50 - 82.76)^2 +...................+ 5 * (100 - 82.76)^2}{91}

\sigma^2 = \frac{11580.6816}{91}

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Lastly, take the square root of the variance to get the standard deviation

\sigma = \sqrt{\sigma^2}

\sigma = \sqrt{127.26}

\sigma = 11.28 --- approximated

<em>Hence, the standard deviation is approximately 11.28</em>

Considering the calculated mean (i.e. 82.76), the standard deviation (i.e. 11.28) is small and this means that the grade of the students are close to the average grade.

8 0
2 years ago
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