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Natali [406]
2 years ago
8

Carolyn has $2.55 in her purse in nickels and dimes. The number of nickels is nine less than three times the number of dimes. Fi

nd the number of each type of coin.
Mathematics
1 answer:
Flauer [41]2 years ago
5 0

Hi, the answer is ;

Answer:

12 dimes and 27 nickels

Step-by-step explanation:

.05N + .1d = 2.55

N= 3d - 9

0.05 (3d-9) + 0.1d = 2.55

0.15d - 0.45 + 0.1d = 2.55

0.25d = 2.55 + 0.45

0.25d = 3

D = 12

N = 3 (12) - 9

N = 36 - 9

N = 27

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A class in advanced physics is composed of 10 juniors, 30 seniors, and 10 graduate students. the final grades show that 3 of the
const2013 [10]

Solution: We are given:

Total students = 10 + 30 + 10 =50

Number of students who received A grade = 3 + 10 + 5 = 18

Number of senior student's who received A grade is = 10

Let A be the event that the student is senior and B be the event that he or she earned an A. Then,

P(A \cap B)=\frac{10}{50}

P(B) = \frac{18}{50}

Now the probability a student chosen at random from this class and is found to have earned an A, the probability that he or she is a senior is:

P(A|B) = \frac{P(A \cap B}{P(B)}

                 =\frac{\frac{10}{50} }{\frac{18}{50} }

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5 0
2 years ago
A runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her
bezimeni [28]
We know, acceleration = change in speed / time
a = (3.5 - 3.1) / 15
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4 0
2 years ago
If mBC = (9x-53) and mCD = (2x + 45) find mBAD
Rom4ik [11]

Answer:

m\angle BAD=73^o

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Find the value of x

Let

O ----> the center of the circle

we know that

Triangle BOC≅Triangle COD

m\angle BOC=arc\ BC ----> by central angle

m\angle COD=arc\ CD ----> by central angle

m\angle BOC=m\angle COD

therefore

arc\ BC=arc\ CD

substitute the given values

(9x-53)^o=(2x+45)^o

solve for x

9x-2x=45+53\\7x=98\\x=14

step 2

Find the measure of angle BAD

we know that

The inscribed angle is half that of the arc it comprises.

so

m\angle BAD=\frac{1}{2} [arc\ BC+arc\ CD]

arc\ BC=9(14)-53=73^o

arc\ CD=2(14)+45=73^o

substitute

m\angle BAD=\frac{1}{2} [73^o+73^o]=73^o

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2 years ago
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96 messages .... $4.80
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