Answer:
The probability that there are 3 or less errors in 100 pages is 0.648.
Step-by-step explanation:
In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.
For the given Poisson distribution the mean is p = 0.03 errors per page.
We have to find the probability that there are three or less errors in n = 100 pages.
Let us denote the number of errors in the book by the variable x.
Since there are on an average 0.03 errors per page we can say that
the expected value is,
= E(x)
= n × p
= 100 × 0.03
= 3
Therefore the we find the probability that there are 3 or less errors on the page as
P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)
Using the formula for Poisson distribution for P(x = X ) = 
Therefore P( X ≤ 3) = 
= 0.05 + 0.15 + 0.224 + 0.224
= 0.648
The probability that there are 3 or less errors in 100 pages is 0.648.
From the information above he makes 15 cups in 5 days .
Answer:
1. 0.18333333333
2. 1.5
3. 0.55555556
4. 2.75
5. 0.81818181818
6. 4.1111111111
Have a nice day, brainliest Please? :)
Answer:
Step-by-step explanation:
x is the length of the side opposite the reference angle of 32, and 10 is the side length adjacent to the reference angle. This satisfies the tangent ratio where
and solving for x:
x = 10tan32
A':(-2,-4)
B':(-7,-6)
C':(-1,-13)
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