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Mademuasel [1]
3 years ago
8

A jar if dimes and quarters is worth $4.55. If there are the same number of dimes as quarters, what is the value of only the dim

es?
PLEASE HELP
Mathematics
1 answer:
Monica [59]3 years ago
8 0
For this case, the first thing we must do is define variables.
 We have then:
 x = number of dimes
 y = number of quarters
 We write the system of equations:
 0.10x + 0.25y = 4.55
 x = y
 Solving the system we have:
 x = 13
 y = 13
 The value of the dimes is:
 (0.10) * (13) = 1.3 $
 Answer:
 
The value of only the dimes is:
 
1.3 $
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To solve this problem you must apply the proccedure below:

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 2. You must make a system of equation, as below:

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 0.20x+0.05y=0.15x15   (ii)

 3. Now, you must find the value of x (the liters of the first canned needed) and y (the liters of the other canned needed):

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3 years ago
Mt. McKinley, in Alaska, is the highest mountain in North America at 20,320
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3 years ago
Eric teaches ceramics in his studio.he estimates that one out of every five people who call for information about a class will s
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For given question,

Eric estimates that one out of every five people who call for information about a class will sign up for the class.

Last week he receive nine calls.

We need to find the probability that four or fewer of the people who called will sign up for a class.

Total number of calls = 9

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Since one out of every five people who call for information about a class will sign up for the class.

the probability of success (p) = 1/5

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and the probability of failure (q) = 1 - p

                                                      = 1 - 0.2

                                                      = 0.8

To find the probability that four or fewer of the people who called will sign up for a class.

So, x would take values 0, 1, 2, 3, 4

Using Binomial principal,

For x = 0,

P(x=0)= ~^9C_0(0.2)^0(0.8)^{9-0}\\\\P(x=0)=0.13422

For x = 1,

P(x=1)= ~^9C_1(0.2)^1(0.8)^{9-1}\\\\P(x=1)=0.30199

For x = 2,

P(x=2)= ~^9C_2(0.2)^2(0.8)^{9-2}\\\\P(x=2)=0.30199

For x = 3,

P(x=3)= ~^9C_3(0.2)^3(0.8)^{9-3}\\\\P(x=3)=0.17616

For x = 4,

P(x=4)= ~^9C_4(0.2)^4(0.8)^{9-4}\\\\P(x=4)=0.06606

So, the required probability would be,

P = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

P = 0.1342 + 0.3020 + 0.3020 + 0.1762 + 0.0661

P = 0.9805

Therefore, the probability that four or fewer of the people who called will sign up for a class = 0.9805

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