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Marrrta [24]
3 years ago
9

Precalculus question help

Mathematics
1 answer:
Fittoniya [83]3 years ago
3 0
4 + 12/13 = 64/13
4 + 19/20 = 99/20
4 + 26/27 = 134/27
4 + 33/34 = 169/34
4 + 40/41 = 204/41

top: 64, 99, 134, 169, 204 
bottom: 13, 20, 27, 34, 41 

top:
35(1) + 29 = 64
35(2) + 29 = 99
35(3) + 29 = 134
35(4) + 29 = 169
35(5) + 29 = 204

bottom:
7(1) + 6 = 13
7(2) + 6 = 20
7(3) + 6 = 27
7(4) + 6 = 34
7(5) + 6 = 41



answer is C.

35n + 29
------------
 7n + 6


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M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

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M = \frac{8*4 + 16*4}{4+4}= 12

b) M = \frac{8*3 + 16*5}{3+5}= 13

c) M = \frac{8*5 + 16*3}{5+3}= 11

Step-by-step explanation:

Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.

a) What is the mean for the combined set if both of the original samples have n=4 scores "

For this case we can use the definition of weighted average given by:

M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

And if we replace the values given we have:

M = \frac{8*4 + 16*4}{4+4}= 12

b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5

Using the definition we have:

M = \frac{8*3 + 16*5}{3+5}= 13

c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3

Using the definition we have:

M = \frac{8*5 + 16*3}{5+3}= 11

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