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Luden [163]
2 years ago
5

Landon babysits and works part time at the water park over the summer. One week, he babysat for 3 hours and worked at the water

park for 10 hours and made $109. The next week he babysat for 8 hours and worked at the water park for 12 hours and made $177. How much does Landon make per hour at each job?
Mathematics
1 answer:
Pavel [41]2 years ago
5 0

Answer:

Babysitting =$8.7

Water park =$8.7

Step-by-step explanation:

In 1 week

Babysitting =3 hrs

Water park = 10 hrs

Total =$ 109

In week 2

Babysitting = 8 hrs

Water park = 12 hrs

Total= $ 177

Both weeks

Babysitting =11 hrs

Water park =22 hrs

Total=$286

33 hrs = $ 286

11 hrs =?

11×286÷33=$95.3

33 hrs = $286

22 hrs = ?

22×286÷33=190.7

Babysitting she got $95.3 in both weeks

Water park she got $190.7(rounded offf)in both weeks

11 hrs = $95.3

1 hr = ?

1×95.3÷11=8.7 (rounded off )

22 hrs = $ 190.7

1 hr = ?

1×190.7÷22= 8.7( rounded off)

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A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?
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Answer:

(a) Selection = 6724520

(b) At\ most\ 12 = 6553976

(c) At\ most\ 8 = 6066720

(d) At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

Step-by-step explanation:

Given

Colors = 8

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Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

n => 28 + 8-1= 35

r = 28

Selection = ^{35}^C_{28

Selection = \frac{35!}{(35 - 28)!28!}

Selection = \frac{35!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29}{7!}

Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}

Selection = \frac{33891580800}{5040}

Selection = 6724520

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

n => 15 + 8 -1 = 22

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At\ least\ 13 = ^{22}C_{15}

At\ least\ 13 = \frac{22!}{(22-15)!15!}

At\ least\ 13 = \frac{22!}{7!15!}

At\ least\ 13 = 170544

Ways of selecting at most 12  =

At\ most\ 12 = Total - At\ least\ 13 --- Complement rule

At\ most\ 12 = 6724520- 170544

At\ most\ 12 = 6553976

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

n => 19+ 8 -1 = 26

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At\ least\ 9 = ^{26}C_{19}

At\ least\ 9 = \frac{26!}{(26-19)!19!}

At\ least\ 9 = \frac{26!}{7!19!}

At\ least\ 9 = 657800

Ways of selecting at most 8  =

At\ most\ 8 = Total - At\ least\ 9 --- Complement rule

At\ most\ 8 = 6724520- 657800

At\ most\ 8 = 6066720

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

n =6+6-1 = 11

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Selection =

^{11}C_6 = \frac{11!}{(11-6)!6!}

^{11}C_6 = \frac{11!}{5!6!}

^{11}C_6 = 462

Using inclusion/exclusion rule of two sets:

Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)

Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462

Only\ 12\ red\ and\ only\ 8\ blue\ = 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  6724520 - 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

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