Answer:
0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand
Step-by-step explanation:
For each customer, there are only two possible outcomes. Either they choose the leading brand, or they do not. The probability of a customer choosing the leading brand is independent of any other customer, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
The leading brand of dishwasher detergent has a 30% market share.
This means that 
A sample of 25 dishwasher detergent customers was taken.
This means that 
a. What is the probability that 10 or fewer customers choose the leading brand?
This is:

In which












Then

0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand
Answer:
First image: -83/2
Second image: 8.62%
Third image: -43/2
Step-by-step explanation:
See the attachments
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The absolute value
returns the "positive version" of a number.
In other words, if the number is positive, it remains positive; if the number is negative, it changes sign.
So, if we want
, we want the "positive version" of x to be 9.
This can happen in two ways: if x is already 9, then its absolute value is still nine. If instead x=-9, its positive value will be 9 again.
In formula, we have

because
