I have a d in my class but i need it caught up to a b or an a asap so will anyone help me please?
2 answers:
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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.
Answer:
b=2/3
Step-by-step explanation:
The table is given as:
The exponential function is given in the form
where a is the initial value and b is the multiplicative rate of change
When x=1, y=9, we have:
When x=3, y=4, we have:
Dividing the two equations:
The multiplicative rate of change, b is 2/3.