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Alja [10]
2 years ago
7

Three friends in Seattle each told you it’s rainy, and each person has a 1/3 probability of lying. What is the probability that

Seattle is rainy? Assume the probability of rain on any given day in Seattle is 0.25.
Mathematics
1 answer:
Sedaia [141]2 years ago
4 0

Answer:

8/11 probability

Step-by-step explanation:

Bayesian stats: you should estimate the prior probability that it’s raining on any given day in Seattle. If you mention this or ask the interviewer will tell you to use 25%. Then it’s straight-forward:

P(raining | Yes,Yes,Yes) = Prior(raining) * P(Yes,Yes,Yes | raining) / P(Yes, Yes, Yes)

P(Yes,Yes,Yes) = P(raining) * P(Yes,Yes,Yes | raining) + P(not-raining) * P(Yes,Yes,Yes | not-raining) = 0.25*(2/3)^3 + 0.75*(1/3)^3 = 0.25*(8/27) + 0.75*(1/27)

P(raining | Yes,Yes,Yes) = 0.25*(8/27) / ( 0.25*8/27 + 0.75*1/27 )

**Bonus points if you notice that you don’t need a calculator since all the 27’s cancel out and you can multiply top and bottom by 4.

P(training | Yes,Yes,Yes) = 8 / ( 8 + 3 ) = 8/11

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at a soccer camp there is a boy to girl ratio of 9 to 7. if there's 63 girls how many boys were there?​
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Answer:

81 boys

Step-by-step explanation:

There are two ways to figure this out. Both ways are correct.

1. Fractions

9/7=x/63

from here you cross multiply

7x=9×63

7x=567

now to get x by itself you divide by 7

x=567÷7

x=81

2. If you have 63 girls and 7 girls to each group. You can divide 63 by 7 and find out how many groups you have which is 9. 9 groups, 7 girls each. For every 7 girls there are 9 boys. 9×9 ( 9 groups, 9 boys each)

9×9=81

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Step-by-step explanation:

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3 years ago
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
Pavel [41]

Answer:

The critical numbers/values are x = 0, 4/7, 2

Step-by-step explanation:

This is a doozy; no wonder you have it up here for help!

The critical numbers of a function are found where the derivative of the function is equal to 0.  To find these numbers, you have to factor the deriative or simply solve it for 0.  This one is especially difficult since it involves rational exponents that have to be factored.  But this is fun, so let's get to it.

First off, I am assuming that the function is

f(x)=x^{\frac{4}{5}}*(x-2)^2 which involves using the product rule to find the derivative.

That derivative is

f'(x)=x^{\frac{4}{5}}*2(x-2)+\frac{4}{5}x^{-\frac{1}{5}}(x-2)^2 which simplifies down to

f'(x)=x^{\frac{4}{5}}(2x^{\frac{5}{5}}-4)+\frac{4}{5}x^{-\frac{1}{5}}(x^{\frac{10}{5}}-4x^{\frac{5}{5}}+4) and

f'(x)=2x^{\frac{9}{5}}-4x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}

Let's get everything over the common denominator of 5 so we can easily add and subtract like terms:

f'(x)=\frac{10}{5}x^{\frac{9}{5}}-\frac{20}{5}x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}

Combining like terms gives us

f'(x)=\frac{14}{5}x^{\frac{9}{5}}-\frac{36}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}

This, however, factors so it is easier to solve for x.  First we will set this equal to 0, then we will factor out

\frac{2}{5}x^{-\frac{1}{5}}:

0=\frac{2}{5}x^{-\frac{1}{5}}(7x^2-18x+8)

By the Zero Product Property, one of those terms has to equal 0 for the whole product to equal 0.  So

\frac{2}{5}x^{-\frac{1}{5}}=0 when x = 0

And

7x^2-18x+8=0 when x = 2 and x = 4/7

Those are the critical numbers/values for that function.  This indicates where there is a max value or a min value.

5 0
3 years ago
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