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garri49 [273]
3 years ago
14

I only need the equations, I can solve it.

Mathematics
2 answers:
xxMikexx [17]3 years ago
6 0
You have a pile of 22 coins that includes nickels and dimes
Let n= nickels and d= dimes
total coins= nickels and dimes
n+d=22
The total value of the coins is $1.20
d=1.20-n
You would solve by system of equations.
n+d=22
d=1.20-n
I suggest using process of substitution
natali 33 [55]3 years ago
4 0
How can you solve it if you don't know the equations?  :P  Just kidding...

n+d=22, so we can say that d=22-n

5n+10d=120, and using d found above in this equation gives you:

5n+10(22-n)=120

5n+220-10n=120

-5n=-100

n=20, and since  d=22-n, d=2

So there are two dimes and twenty nickels...

check...

20(5)+2(10)=100+20=120 cents which is $1.20

n+d=22, 20+2=22, 22=22
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7 0
3 years ago
What is the best approximation of a 15% tip on a $38 car wash?
Neko [114]

Answer:The tip is 5.70 and the whole price is 43.70

Step-by-step explanation:those are exact

3 0
2 years ago
Read 2 more answers
A student has a class with four tests, each worth 100 points, and has earned these
Wewaii [24]

Answer:

average = 52, 100 on fourth test

Step-by-step explanation:

average = 75 + 74 + 71 =  156 / 3 = 52

average = 52

b grade = 75 + 74 + 71 + 100 = 320 / 4 = 80

to get a b grade she must get 100 on her fourth test

4 0
2 years ago
An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a d
anygoal [31]

Answer:

The probability of event <em>A</em> is \frac{7}{11}.

The probability of event <em>B</em> is \frac{6}{11}.

Step-by-step explanation:

The sample space of rolling a fair six-sided dice is as follows:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

      (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

      (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

      (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

      (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

      (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Now consider the experiment of the computing the sum of the two rolls as follows:

X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Number of total outcomes, <em>N</em> = 11.

The probability of an event <em>E</em> is the ratio of the number of favorable outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

The event <em>A</em> is defined as the sum is greater than 5.

The sample space of <em>A</em> is:

A = {6, 7, 8, 9, 10, 11, 12}

n (A) = 7

Compute the probability of event <em>A</em> as follows:

P(A)=\frac{n(A)}{N}=\frac{7}{11}

Thus, the probability of event <em>A</em> is \frac{7}{11}.

The event <em>B</em> is defined as the sum is an even number.

The sample space of <em>B</em> is:

B = {2, 4, 6, 8, 10, 12}

n (B) = 6

Compute the probability of event <em>B</em> as follows:

P(B)=\frac{n(B)}{N}=\frac{6}{11}

Thus, the probability of event <em>B</em> is \frac{6}{11}.

7 0
3 years ago
Which situation would not be represented by the integer -9?
Aneli [31]
The correct answer would be

A football pass of 9 yards
6 0
2 years ago
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