PLEASE HELP DUE VERYSOON
1 answer:
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The probability that the transistor will last between 12 and 24 weeks is 0.424
X= lifetime of the transistor in weeks E(X)= 24 weeks
O,= 12 weeks
The anticipated value, variance, and distribution of the random variable X were all provided to us. Finding the parameters alpha and beta is necessary before we can discover the solutions to the difficulties.
X~gamma()
E(X)=
=
=6 weeks
V(x)=
=24/6= 4
Now we can find the solutions:
The excel formula used to create Figure one is as follows:
=gammadist(X, ,
, False)
P()
P()
P()
P= 0.424
Therefore, probability that the transistor will last between 12 and 24 weeks is 0.424
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Answer:
None of these.
Step-by-step explanation:
Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.
Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.
In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.
If we look at the first term x^3 and we divide it by 13 we get we cannot reduce it so it is not a fraction so 13 is not a factor of P
None of these is the right option.