0.05 kilogramos.................
Based on the absolute deviations and the predicted values, the sum of absolute deviations will be <u>4.8.</u>
<h3>What would be the sum of absolute deviations from predicted values?</h3>
This can be found as:
= ∑ (Observed value - Predicted value)
The observed values are given in the table and the predicted values will be calculated using y = 3.6x - 0.4.
Solving gives:
= [3 - (3.6 x 1 - 0.4)] + [7 - (3.6 x 2 - 0.4)] + [ 9 - (3.6 x 3 - 0.4)] + [14 - (3.6 x 4 - 0.4)] + [15 - (3.6 x 5 - 0.4)] + [21 - (3.6 x 6 - 0.4)] + [25 - (3.6 x 7 - 0.4)]
= 0.2 + 0.2 + 1.4 + 0 + 2.6 + 0.2 + 0.2
= 4.8
Find out more on absolute deviation at brainly.com/question/447169.
Answer:
20π ft³
Step-by-step explanation:
I just did it on the USATestprep
Y=-2/3x+-4. Here is the original equation for a line. (M is slope and b is y-intercept((the point which the line goes through the Y-axis))) y=Mx+b
Below is the solution:
<span>8t + 5p = 220
5t + p = 112
From the second equation:
p = 112 - 5t
Replace into first:
8t + 5(112 - 5t) = 220
t = 20
p = 112 - 5t = 112 - 100 = 12
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