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sergejj [24]
3 years ago
12

Your car insurance comes due annually and generally costs about $1,500. You decide that you would like to set aside a monthly am

ount, beginning in January, to be prepared for when this bill comes at the end of the year. How much should you set aside each month?
Mathematics
1 answer:
Volgvan3 years ago
5 0

Answer:

$125

Step-by-step explanation:

Since there are 12 months in a year, divide the total bill by the total number of months.

1500/12 = $125

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Let width = x, Therefore the length must equal 2x

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3 years ago
Two mechanics worked on a car. The first mechanic worked for
Temka [501]
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x + y = 125

20x + 15y = 2100

Solving:

x = 125 - y

20(125 - y) + 15y = 2100
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During the first stages of an epidemic, the number of sick people increases exponentially with time. Suppose that at = 0 days th
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T = days passed

r = rate of growth

by 0 day, or t = 0, there are 2 folks sick,

\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
P=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &0\\
\end{cases}
\\\\\\
2=P(1+r)^0\implies 2=P\cdot 1\implies 2=P\qquad \boxed{A=2(1+r)^t}

 by the third day, t = 3, there are 40 folks sick,

\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &40\\
P=\textit{initial amount}\to &2\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &3\\
\end{cases}
\\\\\\
40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r
\\\\\\
\sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}

how many folks are there sick by t = 6?   \bf \stackrel{that~many}{A=2(2.7)^6}
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