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.
14,494
To round off the height value to the nearest thousand we can use the
expanded from to clarity the position of numbers which is:
10, 000 = ten thousand
4, 000 = thousands
400 = hundreds
90 = tens
4 = ones
Here we can notice than four thousand is the value where the nearest
thousands is placed. Hence we can round off the number of 14, 494 into 14, 000.
Notice 0-4 rounding off rules.
(5/4)•(4/3)•(3/5)
So you multiply all of them basically
Answer:
I not sure this 76.5
Step-by-step explanation:
9cmx17cm=156/2=76.5
Then 76.5/2= 39
I hope this helpful for you.... Sorry....