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lozanna [386]
3 years ago
7

LOTS OF POINTS HELP

Mathematics
2 answers:
Lesechka [4]3 years ago
8 0

Answer:

Jessica has 48 different sets of classes to choose from.

Step-by-step explanation:

Given

The four categories are:

Economics: 5

Mathematics: 4

Education: 3

Sociology: 2

Suppose that sections for the 3 most popular classes in Economics are full.

Economics is left with (5-3) choices

Economics = 2 choices

Since there are equal chances of making selections, we multiply the number of options in each category to get the total set of classes Jessica can chose from.

Number of choices = 2 * 4 * 3 * 2

Number of choices = 48 choices

Hence, Jessica has 48 different sets of classes to choose from.

Volgvan3 years ago
5 0
<span>Category         Number of Choices
Economics           5 - 3 = 2
Mathematics        4
Education            3
Sociology             2

Since out of 5 sections in Economics 3 are full, there are only 2 remaining sections available for Jessica to choose from.

I think this is a combination: 

n = 2 + 4 + 3 + 2 = 11
r = 1 + 1 + 1 + 1 = 4

nCr = n! / r!(n-r)! = 11! / 4!(11-4)! = 11! / 4! * 7! = 330

Jessica can have 330 different sets of classes to take.

</span>
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notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

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