The answer is 7/8 so you would round it to 1.
Numerator of the equation is 3*nDenominator is 4*p - 5*n The equation is m = 3n/(4p-5n)
Numerator = 3*n
Denominator = 4*p - 5*n
Equation = [m = 3n/(4p-5n)]
The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118 .
In the question ,
it is given that ,
the probability of committing at least , type I error is = 0.7
we have to find the number of tests ,
let the number of test be n ,
the above mentioned situation can be written as
1 - P(no type I error is committed) ≥ P(at least type I error is committed)
which is written as ,
1 - (1 - 0.01)ⁿ ≥ 0.7
-(0.99)ⁿ ≥ 0.7 - 1
(0.99)ⁿ ≤ 0.3
On further simplification ,
we get ,
n ≈ 118 .
Therefore , the number of tests are 118 .
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1 negative startroot 2 end root 1 start root 2 end root
Answer:
Option A= 250
Step-by-step explanation:
Coterminal angles are angles in standard positions having thesame terminal .
Coterminal angles are always negative and positive.
For example:
The coterminal angle of 30° is-
30 - 360 = -330
30 + 360 = 390
Therefore, -330 and 390 are coterminal.
So, to the question:
The angles between 0 -360 coterminal to -110 is 250.
Prove:
250 - 360 = -110
250 + 360 = 610
Option A is correct
