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Anton [14]
2 years ago
11

I need help) Han and Clare go shopping, and they each have a coupon. Answer each question and show your reasoning.

Mathematics
1 answer:
egoroff_w [7]2 years ago
6 0

The amount of money that Han saves by using a coupon is $1.5%

The percentage off Clare's purchase with the coupon is 25%

<h3>How much did Han save?</h3>

A coupon reduces the price at which an item is sold

Amount saved = coupon discouunt x normal price

10% x $15

0.1 x $15 = $1.5

<h3>What is the percent off Clare's coupon?</h3>

Percent off = (6/24 x 100  = 25%

To learn more about how to calculate discounts, please check: brainly.com/question/26061308

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Answer:

34.9^2

Step-by-step explanation:

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we plug 2 in.

pi 4

12.6/2

6.3

6.3 * 3 = 18.9cm^2

18.9 + 16 = 34.9

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Quinzel earns $1254 each month. His total deductions are 20% of his pay. How much is deducted from his pay each month ?
vova2212 [387]

20% = 0.20

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7 0
4 years ago
Could someone help me please!! I'm not getting any of the answers! (Math)
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Read 2 more answers
A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defec
Sphinxa [80]
<span>1.

P(</span>at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

           =n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways, 

where  C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}=  \frac{35*34*33*32}{4!}=  \frac{35*34*33*32}{4*3*2*1}=  52,360

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825

4. so, P(none of the selected calculators is defective)=\frac{52,360}{292,825} =0.18


5. P(at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective)=1-0.18=0.82



Answer:0.82
7 0
4 years ago
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