Step-by-step explanation:
(2x³y^-4)^-2 = 1/(4x⁶y^‐8) = y⁸/(4x⁶)
1/(8x⁵y^-7) = y⁷/(8x⁵)
4x^-9y⁴ = 4y⁴/x⁹
so, the whole expression is then
y⁸/(4x⁶) × y⁷/(8x⁵) × 4y⁴/x⁹
4y¹⁹/(32x²⁰) = y¹⁹/(8x²⁰)
so, D is the correct answer.
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 probability of 3 failures or less is about 0.83886, so that is the probability the machine will be working.
company a: 2mins
company b: 7mins
:)
You're basically told that Lin has traveled 20% of 75. Since 20% is one fifth, she has traveled 75/5=15 miles so far.
