Mean absolute deviation of 75, 89, 145, 85, 80, 92, 104, 90, 100?

1 answer:

7 0

First, compute the mean: $m=\frac{75+89+145+85+80+92+104+90+100}9=\frac{860}9=95.555$

Next compute the absolute deviations for each datapoint:
|75-95.555|=20.555
|89-95.555|=6.555
|145-95.555|=49.445
|85-95.555|=10.555
|80-95.555|=15.555
|92-95.555|=3.555
|104-95.555|=8.445
|90-95.555|=5.555
|100-95.555|=4.445

Next we'll take the mean of these deviations: $\frac{20.555+6.555+49.445+10.555+15.555+3.555+8.445+5.555+4.445}9=\frac{124.665}9=13.8517$

The mean absolute deviation of your data is 13.8517

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20 POINTS! PLS HELP!

Arturiano [62]

First statement is correct. If Sasha reads her novel for 94 minutes, she meets the minimum reading requirement without reading any of the autobiography.

Given,

The rate of reading of a novel by Sasha = 0.8 pages per minute

The rate of reading of autobiography by Sasha = 0.5 pages per minute

The reading requirement of Sasha's English class = atleast 75 pages per week

The given equation: 0.8n + 0.5a ≥ 75

n is the number of minutes she reads the novel

a is the number of minutes she reads the autobiography

Now, let's check the statements:

Statement 1: If Sasha reads her novel for 94 minutes, she meets the minimum reading requirement without reading any of the autobiography.

That is,

0.8 × 94 + 0.5 × 0 ≥ 75

75.2 ≥ 75