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Art [367]
3 years ago
12

Sue has 100 dimes and quarters. If the total value of the coins is $21.40, how many of each kind of coin does she have?

Mathematics
1 answer:
VMariaS [17]3 years ago
8 0
Let's set up the equations we know.
Q+D=100, where Q=number of quarters, and D=number of dimes
.25Q + .1D = 21.40 (because a quarter is worth .25 and a dime worth .10)

Using the first equation, let's isolate one of our variables. I'll pick Q.
Q+D=100
Subtract D from each side
Q=100 - D

Now that we have this new equation, let's plug it into our second equation.
.25Q + .1D = 21.40
.25(100-D) + .1D = 21.40
Distribute the .25
25 - .25D + .1 D = 21.40
Combine like terms
25 - .15D = 21.40
Subtract 21.40 from each side
3.6 - .15D = 0
Add .15 D to each side
3.6 = .15D
Divide each side by .15
D = 24

Now that we have D, we can solve for Q
Q= 100 - D
= 100 - 24
Q = 76

Sue has 24 dimes and 76 quarters.

Let's check our work to make sure.
.25(76) + .1(24) = 
19 + 2.4 = 21.4 √
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Answer:

0.3968 = 39.68% probability this shipment passes inspection.

Step-by-step explanation:

The parts are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question:

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Answer:

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