X(30;0)
y(0;50)
substitue 0 for y and solve for x
substitute 0 in for x and solve for y
Okokokokooo ansnwer is?!??
Answer:
9/16
Step-by-step explanation:
First, note that 10000 multiplied by .5625 will get rid of the decimal point. Therefore, make a fraction where the numerator is .5625 times 10000 and denominator is 10000 like this:
.5625 x 10000
10000
Next, multiply the numbers in the numerator together and keep the denominator as is. Now our fraction looks like this:
5625
/10000
The greatest common factor of 5625 and 10000 is 625, which means you can divide the numerator and denominator by 625 and keep the same value:
5625 ÷ 625
/10000 ÷ 625
And when calculating the numerator and denominator in our fraction above, we get .5625 as a fraction in the simplest form possible:
.5625 =
9
/16
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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