Using the binomial distribution, it is found that there is a:
a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.
b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.
c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.
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For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.
Binomial probability distribution

The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 0.96516 probability of a chipmunk living through the year, thus

Item a:
- Two is P(X = 2) when n = 2, thus:

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.
Item b:
- Six is P(X = 6) when n = 6, then:

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.
Item c:
- At least one not living is:

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.
A similar problem is given at brainly.com/question/24756209
It would be 93.75 pounds as the conversion rate goes for dollar bills than pounds
You could use the solution of 9/4 times 2
A^2 + b^2 = c^2
3^2 + 4^2 = c^2
9 + 16 = c^2
25 = c^2
5 = c
In order for the ball to be used in the game, it must be able to meet the minimum and maximum weight requirements. These are the limits of the weight of the ball. If it exceeds the maximum limit of 16 ounces or below the minimum limit of 14 ounces, the ball will not be approved.
So, by adding 1.5 ounces, that would mean that the initial weight of the ball did not reach the minimum limit. The initial weight of the ball, denoted as x, have two possible values. The first value is the initial weight plus added with 1.5 ounces would reach 14 ounces.
x + 1.5 = 14
x = 12.5 ounces
The other scenario is when the initial weight is added to reach the maximum requirement of 16 ounces
x + 1.5 = 16
x = 14.5 ounces
From both answer, we could conclude that the initial weight has to be 12.5 ounces. If the initial weight were 14.5 ounces to begin with, there should be no need for air. It cans till be approved to be used.