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stiv31 [10]
1 year ago
7

You found $6.60 on the ground at school, all in nickels, dimes, and quarters. you have twice as many quarters as dimes and 42 co

ins in all. howmany of each type of coin do you have
Mathematics
1 answer:
Veseljchak [2.6K]1 year ago
8 0
Let the number of nickels found be n, the number of dimes be d and the number of quarters be q.

i) "you have twice as many quarters as dimes and 42 coins in all."

means that 2d=q, and n+d+q=42

we can reduce the number of unknowns by substituting q with 2d:

n+d+q=42
n+d+2d=42
n+3d=42

We can write all n, d and q in terms of d as follows:

there are n=42-3d nickels, d dimes and q=2d quarters.

ii) In total there are $6.60 dollars,

1 nickel     =  5 cent   =  $0.05
1 dime      =  10 cent  =  $0.1
1 quarter   =  25 cent  = $0.25 

thus

(42-3d)*0.05 + d*0.1 +2d*0.25= $6.60

2.1 - 0.15d+0.1d+0.5d=6.60

2.1+0.45d=6.6

0.45d=6.6-2.1=4.5

d=4.5/0.45=10

iii)

so, there are 10 dimes, 2d=2*10=20 quarters and 42-3d=42-3*10=12 nickels.


Answer: 10 dimes, 20 quarters, 12 nickels
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Step-by-step explanation:

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2 years ago
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A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if
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Answer:

a) The expected value is \frac{-1}{15}

b) The variance is  \frac{49}{45}

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is \binom{10}{2}.

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is \binom{5}{2}. Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

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Let X be the expected value of the amount you can win. Then,

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We will use the following formula

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Thus

Var(X) = \frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}

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