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Trava [24]
3 years ago
12

Each month cameron budgets $1040 for fixed expenses $980 for living expenses and $120 for annual expenses his annual net income

is $24,132. Which statement best describes his monthly budget?
A. It shows a deficit of $129
B. it shows a surpus of $129
C. It shows a deficit of $9
D. it is Balanced
Mathematics
2 answers:
sineoko [7]3 years ago
5 0
Let us first find how much he is spending every year, to do this let's find his monthly expense and multiply it by 12.


Every month Cameron spends 1040 + 980 + 120 = 2140

To find out how much he spends yearly, multiply the monthly value by 12,
2140 x 12 = 25680


This value is more than his net income so he clearly has a surplus, but to check we can subtract 129 from every month to get:

(1040 + 980 + 120 - 129) = 2011

2011 x 12 = 24132, which shows that his budget is saving 129 surplus every month. Choice B is correct.
shtirl [24]3 years ago
5 0

Answer: Option 'A' is correct.

Step-by-step explanation:

Since we have given that

Fixed expenses = $1040

Living expenses = $980

Annual expenses = $120

So, Total expenditure would be

\$1040+\$980+\$120\\\\=\$2140

Annual net income = $24,132

Monthly income(receipt) would be

\dfrac{24132}{12}=\$2011

Since we can see that total expenditure is more than total receipts.

So, there will be deficit.

Deficit = Total expenditure - Total receipt

Deficit = $2140-$2011

Deficit = $129

Hence, there is a deficit of $129.

Thus, option 'A' is correct.

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Answer:

This is the rate at which the radius of the balloon is changing when the volume is 300 ft^3 \frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \:\frac{ft}{h}  \approx -0.05537 \:\frac{ft}{h}

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V=\frac{4}{3}\pi r^3\\\\300=\frac{4}{3}\pi r^3\\\\r^3=\frac{225}{\pi }\\\\r=\sqrt[3]{\frac{225}{\pi }}

Substitute the values we know and solve for \frac{dr}{dt}

\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=-\frac{12}{4\pi (\sqrt[3]{\frac{225}{\pi }})^2} \\\\\frac{dr}{dt}=-\frac{3}{\pi \left(\sqrt[3]{\frac{225}{\pi }}\right)^2}\\\\\frac{dr}{dt}=-\frac{3}{\pi \frac{225^{\frac{2}{3}}}{\pi ^{\frac{2}{3}}}}\\\\\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \approx -0.05537 \:\frac{ft}{h}

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