Answer:
30,60,90 triangle, which is the lengths 1 √
3
, and 2
respectively. This is what I have got so far: Using the 30-60-90 Ratio
Step-by-step explanation:
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= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
Step-by-step explanation:
for
I= ∫x^n . e^ax dx
then using integration by parts we can define u and dv such that
I= ∫(x^n) . (e^ax dx) = ∫u . dv
where
u= x^n → du = n*x^(n-1) dx
dv= e^ax dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)
then we know that
I= ∫u . dv = u*v - ∫v . du + C
( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =
(u*v) - ∫v*du + C )
therefore
I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
12 laps because Lee ran 3 and that's half of what ray ran so times that by 2 and then times 6 by 2 and that's your answer
Answer:
any one need a gf??
Step-by-step explanation: HELLO here she is waiting for u may decline but hey worth a shot.... And id.k the answer
