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Valentin [98]
3 years ago
7

Nina heard that as a general rule, she should spend no more than one week's pay on rent. If Nina's pay is $27,600 per year, what

is the maximum amount per month that she should spend on rent?
Its $530!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Mathematics
2 answers:
SIZIF [17.4K]3 years ago
8 0

52 weeks in a year

27600/52 = 530.769

so 1 weeks pay = 530.77

 she shouldn't pay more than that on rent


(Round answer as needed)

 

Hoochie [10]3 years ago
5 0

Answer:

\$530.77

Step-by-step explanation:

we know that

1\ year=52\ weeks

step 1

Divide the total pay in a year by the total weeks in a year

so

\frac{\$27,600}{52}= \$530.77 ----> one week's pay

therefore

The maximum amount per month that she should spend on rent is \$530.77

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Quadrilateral LMNO is similar to quadrilateral PQRS. Find the measure of side QR.
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The measure of the side QR is 8.8.

Step-by-step explanation:

Since LMNO \sim PQRS, then SP \propto OL, RS \propto ON, RQ \propto MN and QP \propto LM. From figure we have the following relationship:

k = \frac{OL}{SP} = \frac{ON}{RS} = \frac{MN}{QR} = \frac{ML}{QP} (1)

Where k is the proportionality ratio.

If we know that SP = 13, OL = 55, MN = 37, then the measure of side QR is:

k = \frac{OL}{SP} (1b)

k = \frac{55}{13}

k = \frac{MN}{QR}

QR = \frac{MN}{k} (1c)

QR = \frac{37}{\frac{55}{13} }

QR = 8.745

The measure of the side QR is 8.8.

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A piggy bank contains pennies, nickels, and dimes. The number of dimes is 15 more than the number of nickels, and there are 140
amm1812

Answer:

Option D is correct.

There are 34 nickels in the piggy bank.

Step-by-step explanation:

A piggy bank contains pennies, nickels and dimes.

Let the number of pennies be p

Let the number of nickels be n

Let the number of dimes be d

Also, note that 1 penny = $0.01

1 nickel = $0.05

1 dime = $0.10

- The number of dimes is 15 more than the number of nickels.

d = 15 + n

- There are 140 coins altogether totaling $7.17.

p + n + d = 140

0.01p + 0.05n + 0.1d = 7.17

Bringing the 3 equations together

d = 15 + n (eqn 1)

p + n + d = 140 (eqn 2)

0.01p + 0.05n + 0.1d = 7.17 (eqn 3)

Substitute (eqn 1) into (eqn 2)

p + n + d = 140

p + n + (15 + n) = 140

p + 2n + 15 = 140

p = 140 - 15 - 2n = 125 - 2n

p = 125 - 2n (eqn 4)

Substitute (eqn 1) and (eqn 4) into (eqn 3)

0.01p + 0.05n + 0.1d = 7.17

0.01(125 - 2n) + 0.05n + 0.1(15 + n) = 71.7

1.25 - 0.02n + 0.05n + 1.5 + 0.1n = 7.17

0.1n + 0.05n - 0.02n + 1.5 + 1.25 = 7.17

0.13n + 2.75 = 7.17

0.13n = 7.17 - 2.75 = 4.42

0.13n = 4.42

n = (4.42/0.13) = 34

d = 15 + n = 15 + 34 = 49

p = 125 -2n = 125 - (2×34) = 125 - 68 = 57

Hence, there are 57 pennies, 34 nickels and 49 dimes in the piggy bank.

Hope this Helps!!!

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