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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The fourth one, (3,-6), (3,-4), (6,-4), (6,-6)
When roots of polynomials occur in radical form, they occur as two conjugates.
That is,
The conjugate of (a + √b) is (a - √b) and vice versa.
To show that the given conjugates come from a polynomial, we should create the polynomial from the given factors.
The first factor is x - (a + √b).
The second factor is x - (a - √b).
The polynomial is
f(x) = [x - (a + √b)]*[x - (a - √b)]
= x² - x(a - √b) - x(a + √b) + (a + √b)(a - √b)
= x² - 2ax + x√b - x√b + a² - b
= x² - 2ax + a² - b
This is a quadratic polynomial, as expected.
If you solve the quadratic equation x² - 2ax + a² - b = 0 with the quadratic formula, it should yield the pair of conjugate radical roots.
x = (1/2) [ 2a +/- √(4a² - 4(a² - b)]
= a +/- (1/2)*√(4b)
= a +/- √b
x = a + √b, or x = a - √b, as expected.
