Assume that the total overhead variance is x
We are given that the total labor variance is twice the total overhead variance. This means that, the total labor variance is 2x
Total variance can be calculated as follows:
Total variance = Total materials variance + Total overhead variance
+ Total labor variance
We have:
Total variance = $35000
Total materials variance = $14000
Total overhead variance = x
Total labor variance = 2x
Substitute in the equation and solve for x as follows:
35000 = 14000 + x + 2x
35000 - 14000 = 3x
21000 = 3x
x = 21000/3
x = 7000
Based on the above calculations:
Total overhead variance = x = $7000
Total labor variance = 2x = 2*7000 = $14000
18:20
Multiply each side by 2
gas mileage = k*s, where k is a constant of proportionality and s is the speed. Unfortunately, this does not take into account the fact that the engine consumes fuel even when the car is not moving.
Here it makes most sense to regard {0, S} as the domain for this function. Here, S would represent the car's top speed.
Answer:
91.125
Step-by-step explanation:
Total number/ Total number of quantity
16+32+10+28+8+19+32+26+33+39+98+399
729
Divode by number of quantity (8)
729/8
=91.125
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Answer: 0.0035
Step-by-step explanation:
Given : The readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees C.
i.e. 
and 
Let x denotes the readings on thermometers.
Then, the probability that a randomly selected thermometer reads greater than 2.17 will be :_
![P(X>2.7)=1-P(\xleq2.7)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{2.7-0}{1})\\\\=1-P(z\leq2.7)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9965\ \ [\text{By z-table}]\ \\\\=0.0035](https://tex.z-dn.net/?f=P%28X%3E2.7%29%3D1-P%28%5Cxleq2.7%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B2.7-0%7D%7B1%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq2.7%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.9965%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%20%5C%5C%5C%5C%3D0.0035)
Hence, the probability that a randomly selected thermometer reads greater than 2.17 = 0.0035
The required region is attached below .
