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dybincka [34]
3 years ago
13

A pile of coins, consisting of quarters and half dollars, it worth $11.75. If there are 2 more quarters than half dollars, how m

any of each are there?
Mathematics
1 answer:
lord [1]3 years ago
3 0
Let the number of half dollars be x,

number of quarters = x + 2
amount of half dollars = 0.5x
amount of quarters = 0.25(x + 2)
= 0.25x + 0.5
total amount = 0.5x + 0.25x + 0.5
= 0.75x + 0.5
0.75x + 0.5 = 11.75
0.75x = 11.25
x = 15
number of quarters = x + 2
= 15 + 2
= 17

There are 17 quarters and 15 half dollars.
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MatroZZZ [7]

Answer:

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Step-by-step explanation:

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8 0
2 years ago
An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.
andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

P(A \text{wins})=0.5833

6 0
3 years ago
12÷4×32+(4−2)5
Taya2010 [7]
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7 0
2 years ago
Adrian are 3 ani, iar tatăl lui are 34 de ani. Peste câți ani Adrian va fi de două ori mai tânăr decât tatăl său?
katrin [286]

Answer:

27 years

Step-by-step explanation:

Given that :

Adrian's age = 3

Father's age = 34

In how many years will Adrian be twice as young as his father?

Let the number of years = x

2(3 + x) = 34 + x

6 + 2x = 34 + x

2x - x = 34 - 6

x = 28

3 0
3 years ago
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Tamiku [17]
$20/5 is $4, so for every increase in temperature, it's 4 (1°=$4). So 9*4=36. So your answer is B.
4 0
3 years ago
Read 2 more answers
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