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2 years ago

A car purchased for $24,000 is expected to lose value, or depreciate, at a rate of 8% per year. This situation can be modeled by

P(t)=24,000(1-0.08)^t
What is the meaning of the output P(t) in context of the problem?
What is the meaning of the input t in context of the problem?

Mathematics

1 answer:

algol [13]2 years ago

5 0

$\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &24000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ t=\textit{elapsed time}\dotfill &t\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current value}}{\stackrel{A}{\stackrel{\downarrow }{P(t)}}}=24000(1-0.08)^{\stackrel{\stackrel{\textit{years}}{\downarrow }}{t}}$

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k0ka [10]

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3 years ago

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kupik [55]

Answer:

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Step-by-step explanation:

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We can see that in our figure there are 100 squares as there are 10 rows and each row contains 10 squares.

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To convert the shaded area into percent, we need to multiply 0.81 by 100.

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$\text{Shaded area as percent of the whole}=81\%$

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3 years ago

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A book sold 33,100 copies in its first month of release. Suppose this represents 7.8% of the number of copies sold to date. How

matrenka [14]

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2 years ago

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ycow [4]

Answer:

1/2

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2 years ago

What is the radius of a sphere if its surface area is 2,122.64 square inches? (Use 3.14 for π.) 11 in. 13 in. 121 in. 169 in.

crimeas [40]

To solve this, we are going to use the surface are of a sphere formula: $A=4 \pi r^2$
where
$A$ is the surface area of the sphere
$r$ is the radius of the sphere

We know for our problem that $A=2122.64$ and $\pi =3.14$ , so lets replace those vales in our formula:
$A=4 \pi r^2$
$2122.64=4(3.14)r^2$
$2122.64=12.56r^2$
Now, we just need to solve our equation for $r$ :
$\frac{2122.64}{12.56} =r^2$
$r^2=\frac{2122.64}{12.56}$
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$r=+or- \sqrt{169}$
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Since the radius of a sphere cannot be a negative number, $r=13$ .

We can conclude that the radius of a sphere with surface area <span>2,122.64 </span> $in^2$ is 13 in.

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