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xxMikexx [17]
2 years ago
14

On a shelf of a book store, there are 3 books on English, 4 books on Economics and 5 books on Business Mathematics. How many col

lections can be made, if each collection consists of at most 3 books on each subject?​
Mathematics
1 answer:
shutvik [7]2 years ago
3 0

If each collection consists of at most 3 books on each subject, the number of collection that can be made of 3 English, 4 Economics, and 5 Business Mathematics books are <u>40</u>.

<h3>What is a combination?</h3>

A combination, as a mathematical technique, determines the number of possible arrangements in a collection of items.

In the combination, the order of the selection does not matter.

The implication is that in combinations, one can select the items in any order, unlike permutations, where the order of selection matters.

<h3>Data and Calculations:</h3>

English books on shelf = 3

Economics books on shelf = 4

Business Mathematics books on shelf = 5

Collections to be made = 3 books on each subject

The solution can be set up as a combination problem, as follows:

3 C 3 x 4 C 3 x 5 C 3

3! / 3! (3 – 3)! The solution is 1.

4! / 3! (4 – 3)! The solution is 4.

5! / 3! (5 – 3)! The solution is 10.

1 x 4 x 10

= 40

Thus, if each collection consists of at most 3 books on each subject, the number of collections that can be made of 3 English, 4 Economics, and 5 Business Mathematics books is <u>40</u>.

Learn more about combinations and permutations at brainly.com/question/4658834

#SPJ1

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3 years ago
Teresa bought 16 roses for $20.64. How much did she pay for each rose
jeka94
Roses: 16
Total Cost: $20.64

Divide:
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3 years ago
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Fudgin [204]
You can’t solve this, not enough information
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3 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

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3 years ago
How do you factor out the coefficient of 2.2x+4.4
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3 years ago
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