The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability that there are 3 or less errors in 100 pages? Round your answer to four decimal places (e. g. 98.7654).

Therefore the required probability is 0.6472.

Step-by-step explanation:

Poisson distribution: A Poisson distribution is discrete distribution.

Let X be a discrete variable the number of event. Let λ be the mean of X.

Then the probability of k event is

Here mean of each page is =0.03

Mean of 100 pages = (0.03×100)= 3

λ = 3

The required probability

P(X≤3)  

= P(X=0)+P(X=1)+P(X=2)+P(X=3)

≈ 0.6472

Therefore the required probability is 0.6472

d

step-by-step explanation:

option: d is the correct answer.

d) none of the above.

step-by-step explanation:

clearly from the figure attached to the question we could see that there is no shape given. hence without any shape given in the figure and after rotation we can't obtained a square pyramid.

since, to obtain a image we need a proper figure of the pre-image so that it can be transformed.

hence, the correct answer is:

d) none of the above.

step-by-step explanation: