We first calculate the z-score corresponding to x = 1075 kWh. Given the mean of 1050 kWh, SD of 218 kWh, and sample size of n = 50, the formula for z is:
z = (x - mean) / (SD/sqrt(n)) = (1075 - 1050) / (218/sqrt(50)) = 0.81
From a z-table, the probability that z > 0.81 is 0.2090. Therefore, the probability that the mean of the 50 households is > 1075 kWh is 0.2090.
Answer is D (28)
28°-13°=15°
So it need to rise 28°
To calculate the average you add up all the scores and divide by the number of scores (5)
add all the current test scores and allow X to be the unknown score:
92+80+88+93+X: or 353 + X
set up equation:
(353+X)
______ = 90
5
Multiply both side of the equation by 5 to clear the fraction resulting in:
353+X=450
Subtract 353 from both sides of the equation:
x = 97
So the test score needed to get a 90 average is 97.
M > 6 Also, dont forget to put a line under the arrow or it will be wrong.
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