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guapka [62]
3 years ago
8

A student expressed the sum of 2 whole numbers as 5 times (8+3).What are the 2 whole numbers?

Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0

Well if the two numbers are equal to 5x(8+3).

Lets say that these two numbers are X and Y,

So that would be X + Y = 5(8+3)

which means X + Y = 55

You could write that as a fuctions Y = 55 - X

Now by using the graph you get and the X values of [0,55] ( This means every value from 0 to 55 ), you will get a X and Y value that if you added up together will give you the value of 8(5+33)

An examples for the answers would be

(0,55) (1,54) (2,53) (3,52) (4,51) (5,50) (6,49) ... it goes on until x reaches 55.

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A computer programming team has 13 members. a. How many ways can a group of seven be chosen to work on a project? b. Suppose sev
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Answer:

1716 ;

700 ;

1715 ;

658 ;

1254 ;

792

Step-by-step explanation:

Given that :

Number of members (n) = 13

a. How many ways can a group of seven be chosen to work on a project?

13C7:

Recall :

nCr = n! ÷ (n-r)! r!

13C7 = 13! ÷ (13 - 7)!7!

= 13! ÷ 6! 7!

(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)

1235520 / 720

= 1716

b. Suppose seven team members are women and six are men.

Men = 6 ; women = 7

(i) How many groups of seven can be chosen that contain four women and three men?

(7C4) * (6C3)

Using calculator :

7C4 = 35

6C3 = 20

(35 * 20) = 700

(ii) How many groups of seven can be chosen that contain at least one man?

13C7 - 7C7

7C7 = only women

13C7 = 1716

7C7 = 1

1716 - 1 = 1715

(iii) How many groups of seven can be chosen that contain at most three women?

(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)

Using calculator :

(15 * 35) + (6 * 21) + (1 * 7)

525 + 126 + 7

= 658

c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?

(First in second out) + (second in first out) + (both out)

13 - 2 = 11

11C6 + 11C6 + 11C7

Using calculator :

462 + 462 + 330

= 1254

d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

Number of ways with both in the group = 11C5

Number of ways with both out of the group = 11C7

11C5 + 11C7

462 + 330

= 792

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