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katovenus [111]
3 years ago
11

In 1908, a Model T cost $950. By 1927, Ford had become so efficient that the car's price dropped to $280. What was the percentag

e decrease in price from 1908 to 1927?
Mathematics
1 answer:
topjm [15]3 years ago
8 0

Answer:

Step-by-step explanation:

In 1908, a Model T cost $950. This means that the initial price was $950. The car's price dropped to $280 in the year 1927 due to it's increased efficiency.

The decrease in price is determined by subtracting the current price in 1927 from the initial price in 1908. It becomes

950 - 280 = $670

To determine the percentage decrease in price from 1908 to 1927, we will divide the decrease in price from 1908 to 1927 by the initial price in 1908 and multiply by 100. It becomes

670/950 × 100

= 70.53%

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What is the equation that represents the sequence in this problem ? Find the price after the 8th month
AVprozaik [17]

ANSWER

\begin{gathered} a_n=ar^{n\text{ - 1}} \\ a_8\text{ = \$38.26} \end{gathered}

EXPLANATION

The problem represents a geometric progression.

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a_n=ar^{n\text{ - 1}}

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The first term from the table is the first price (for the first month). That is $80.00

To find the common ratio, we divide a term by its preceeding term.

Let us divide the price of the second month from the first.

We have:

\begin{gathered} r\text{ = }\frac{72}{80} \\ r\text{ = 0.9} \end{gathered}

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So, we have that:

\begin{gathered} a_8\text{ = 80 }\cdot0.9^{(8\text{ - 1)}} \\ a_8\text{ = 80 }\cdot0.9^7 \\ a_8\text{ = \$38.26} \end{gathered}

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8 months ago
Airen’s grandparents deposited $1300 in a mutual fund earning 6% interest compounded annually. Write an equation to represent ho
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~~~~~~ \textit{Compound Interest Earned Amount \underline{in 18 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1300\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &18 \end{cases}

A=1300\left(1+\frac{0.06}{1}\right)^{1\cdot 18}\implies A=1300(1.06)^{18}\implies A\approx 3710.64 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = 1300(1.06)^x~\hfill

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