2 goes into all the terms so;
2(-2x² + 7x - 3) now you could factor it:
2(-2x - 1)(x + 3)
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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I'm not sure what your variables are, but I will try task 3 with you.
y=$500(in billions)
x= the number of years that have passed since 2009
percentage of decrease: 1 - % =
so your formula will look like this: y(1 - %)^x
