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zimovet [89]
3 years ago
8

Will someone help I'm up early trying to do this

Mathematics
2 answers:
zimovet [89]3 years ago
7 0
D. 3g -2/3 Have a good day :D


Allisa [31]3 years ago
3 0
1/3*(9g-2)= 1/3*9g - 2/3 = 3g-2/3

so D is the right answer
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ASAP<br><br> (only the answer pls)
atroni [7]

Answer:

B..........................................

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2 years ago
What is 70/6 as a mixed number
bija089 [108]
70/6 is basically saying "divide 70 by 6". So to get to a mixed numer, thats what you have to do.

70/6=11.666666...
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The answer is 11 2/3.


hope i helped!! :))
6 0
3 years ago
Read 2 more answers
After 8 points are added to each score in a sample, the mean is found to be M= 40. What was the value for the original mean
MArishka [77]

Answer:

The value for the original mean = 32

Step-by-step explanation:

Here, we want to calculate the original value of the mean.

Let the number of samples be n

Mathematically;

mean = Total value/n

Now, we added 8 to each of values; total value added = 8 * n = 8n

Now, for the new mean of 40; we have

(Total value + 8n)/n = 40

Total value + 8n = 40n

Total value = 40n -8n

Total value = 32n

kindly recall from the beginning of the solution;

mean = Total value/n

mean = 32n/n

mean = 32

So the original value of the mean is 32

6 0
2 years ago
Whats the 11th term of 2,6,18,54
Natalija [7]
The 11th term would be 39,366.

The rule for this sequence is 3n (n times 3). Simplying multiplying by 3 until you get to the 11th term (54 * 3 * 3 * 3 * 3 * 3 * 3), would give you this answer.

I hope this helps!!
8 0
3 years ago
Given the volume of Figure A is 512cm ^3and Figure B is 343cm^3, find the ratio of the perimeter from Figure A to Figure B.
sp2606 [1]

\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\[-0.35em] ~\dotfill

\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{\textit{figure A}}{\textit{figure B}}\qquad \qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{512}}{\sqrt[3]{343}}\qquad \begin{cases} 512=&2^9\\ &2^{3\cdot 3}\\ &(2^3)^3\\ 343=&7^3 \end{cases}\implies \cfrac{s}{s}=\cfrac{\sqrt[3]{(2^3)^3}}{\sqrt[3]{7^3}} \\\\\\ \cfrac{s}{s}=\cfrac{2^3}{7}\implies \cfrac{s}{s}=\cfrac{8}{7}\implies s:s = 8:7\impliedby \textit{ratio of the }\stackrel{sides~and}{perimeters}

5 0
3 years ago
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