Answer:1.799908×10^1
Step-by-step explanation:
18-0.00092
=17.99908
In standard form=1.799908×10^1
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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Answer:
I think it is D
Step-by-step explanation:
Hey mate. Here is your answer.
Set up the composite function and evaluate.
g (17x^2 - 10x) = 153x^2 - 90x - 9
Hope this helps.
Answer:
I couldnt write it on the computer because i wanted to draw other stuff to help you understand so i did it on paper. The photo is below!
Step-by-step explanation:
