EXTREMELY URGENT PLZ ANSWER PLZ PLZ
1 answer:
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Answer:
57.5
Step-by-step explanation:
The expected frequency of West Campus and Failed :
Let :
Failed = F
East Campus = C
West Campus = W
Passed = P
Frequency of FnW :
[(FnE) + (FnW) * (PnW) + (FnW)] / total samples
[(52 + 63) * (63 + 37)] / 200
[(115 * 100)] / 200
11500 / 200
= 57.5
n(Failed n East campus)
Answer:
log₂(3) = 1.585 ≠ 1.5
Her thinking is not valid because the technique of average is valid only if the graph of the function is a straight line, but the graph of the log function is not a straight line.
Therefore the values cannot be taken by average
Step-by-step explanation:
Given:
log₂(2) = 1
log₂(4) = 2
To evaluate : log₂(3)
Now,
we know that
logₓ(y) = (Here the log has same base in the numerator and the denominator i.e 10)
therefore,
log₂(3) =
also,
log(2) = 0.3010
log(3) = 0.4771
thus,
log₂(3) =
or
log₂(3) = 1.585 ≠ 1.5
Her thinking is not valid because the technique of average is valid only if the graph of the function is a straight line, but the graph of the log function is not a straight line.
Therefore the values cannot be taken by average