2. Mr. Ellis runs an after-school program for nine- and ten-year-olds. Each day the children participate in an
1 answer:
Answer:
See explanation
Step-by-step explanation:
Factorize numbers 42 and 56:
These two numbers have common factors 2 and 7. So,
A. Mr. Ellis can divide the group into
- 1 team = 42 ten-year-olds and 56 nine-year-olds (actually this is not dividing only completing 1 team);
- 2 teams = 21 ten=year-olds and 28 nine-year-olds in each team;
- 7 teams = 6 ten-year-olds and 8 nine-year-olds in each team;
- 14 teams = 3 ten-year-olds and 4 nine-year-olds in each team.
So, there are 3 different ways to divide the group of students into teams.
B. The greatest number of teams Mr. Ellis can make so each team has the same number of 9-year-olds and the same number of 10-year-olds is 14 teams.
C. If Mr. Ellis gives a snack to each winner, then he is interested to give the smallest number of snacks, the smallest number of snacks will be when the number of students in the team is the smallest, the smallest number of students will be when the greatest number of teams are created.
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Answer:
(a) The probability that the number of successes is at most 5 is 0.1379.
(b) The probability that the number of successes is at most 5 is 0.1379.
(c) The probability that the number of successes is at most 5 is 0.1379.
(d) The probability that the number of successes is at most 11 is 0.9357.
→ All the exact probabilities are more than the approximated probability.
Step-by-step explanation:
Let <em>S</em> = a basketball player scores a shot.
The probability that a basketball player scores a shot is, P (S) = <em>p</em> = 0.30.
The number of sample selected is, <em>n</em> = 25.
The random variable
According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion () follows a normal distribution.
The mean of the the sampling distribution of the sample proportion is:
The standard deviation of the the sampling distribution of the sample proportion is:
(a)
Compute the probability that the number of successes is at most 5 as follows:
The probability of 5 successes is:
**Use the standard normal table for probability.
Thus, the probability that the number of successes is at most 5 is 0.1379.
The exact probability that the number of successes is at most 5 is:
The exact probability is more than the approximated probability.
(b)
Compute the probability that the number of successes is at most 7 as follows:
The probability of 5 successes is:
**Use the standard normal table for probability.
Thus, the probability that the number of successes is at most 7 is 0.4129.
The exact probability that the number of successes is at most 7 is:
The exact probability is more than the approximated probability.
(c)
Compute the probability that the number of successes is at most 9 as follows:
The probability of 5 successes is:
**Use the standard normal table for probability.
Thus, the probability that the number of successes is at most 9 is 0.7422.
The exact probability that the number of successes is at most 9 is:
The exact probability is more than the approximated probability.
(d)
Compute the probability that the number of successes is at most 11 as follows:
The probability of 5 successes is:
**Use the standard normal table for probability.
Thus, the probability that the number of successes is at most 11 is 0.9357.
The exact probability that the number of successes is at most 11 is:
The exact probability is more than the approximated probability.
