Round 4,070 to the nearest thousand
2 answers:
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Answer:
There are 35 different sets of 3 books Joe could choose
Step-by-step explanation:
* Lets explain how to solve the problem
- Combination is a collection of the objects where the order doesn't
matter
- The formula for the number of possible combinations of r objects from
a set of n objects is nCr = n!/r!(n-r)!
- n! = n(n - 1)(n - 2)................. × 1
Lets solve the problem
- Joe has one book each for algebra, geometry, history, psychology,
Spanish, English and Physics in his locker
∴ He has <em>seven</em> books in the locker
- He wants to chose <em>three</em> of them
∵ The order is not important when he chose the books
∴ We will use the combination <em>nCr</em> to find how many different sets
of three books he can choose
- The total number of books is 7
∴ n = 7
∵ He chooses 3 of them
∴ r = 3
∵ 7C3 = 7!/3!(7 - 3)! = 7!/3!(4!)
∴
∴ 7C3 = 35
* There are 35 different sets of 3 books Joe could choose
Answer:
0.3108 is the probability that the sample mean is between 7.8 and 8.2 minutes.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 8 minutes
Standard Deviation, σ = 2.5 minutes
Sample size, n = 25
We are given that the distribution of time spent is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =
P(sample mean is between 7.8 and 8.2 minutes)
0.3108 is the probability that the sample mean is between 7.8 and 8.2 minutes.