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Miguel invested [tex]$\$[/tex] 400[tex]$ in a bank account that pays interest. The account's value over time, in $[/tex]x[tex]$ years, is given in the table.

\begin{tabular}{|c|c|}
\hline
\begin{tabular}{c}
Time \\ (years)
\end{tabular} & \begin{tabular}{c}
Account Value \\ (\$[/tex])
\end{tabular} \\
\hline
0 & 400 \\
\hline
1 & 436 \\
\hline
2 & 468 \\
\hline
3 & 502 \\
\hline
4 & 550 \\
\hline
5 & 589 \\
\hline
\end{tabular}

Which exponential function models the data? Round the numerical values to the nearest hundredth.

A. [tex]$f(x)=1.41(30.69)^x$[/tex]
B. [tex]$f(x)=37.74(396.48)^x$[/tex]
C. [tex]$f(x)=390.60(1.41)^x$[/tex]
D. [tex]$f(x)=401.19(1.08)^x$[/tex]

Sagot :

GinnyAnswer
To determine which exponential function models the data given in the table, let's examine the account values over the time period and compare them to the provided exponential function candidates.

We are given the following data points for the account value over time:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (years)} & \text{Account Value (\$)} \\ \hline 0 & 400 \\ \hline 1 & 436 \\ \hline 2 & 468 \\ \hline 3 & 502 \\ \hline 4 & 550 \\ \hline 5 & 589 \\ \hline \end{array} \][/tex]

We'll compare these data points to each candidate exponential function to see which one fits:

1. [tex]\( f(x) = 1.41 \cdot (30.69)^x \)[/tex]
2. [tex]\( f(x) = 37.74 \cdot (396.48)^x \)[/tex]
3. [tex]\( f(x) = 390.60 \cdot (1.41)^x \)[/tex]
4. [tex]\( f(x) = 401.19 \cdot (1.08)^x \)[/tex]

For each candidate function, we substitute [tex]\( x \)[/tex] with the time values from 0 to 5 and compare the resulting values to the account values given.

Let's check if these functions match the data points:

- [tex]\( f(x) = 1.41 \cdot (30.69)^x \)[/tex]
- [tex]\( f(0) = 1.41 \cdot (30.69)^0 = 1.41 \)[/tex]
- This does not match the initial account value of 400.

- [tex]\( f(x) = 37.74 \cdot (396.48)^x \)[/tex]
- [tex]\( f(0) = 37.74 \cdot (396.48)^0 = 37.74 \)[/tex]
- This does not match the initial account value of 400.

- [tex]\( f(x) = 390.60 \cdot (1.41)^x \)[/tex]
- [tex]\( f(0) = 390.60 \cdot (1.41)^0 = 390.60 \)[/tex]
- This is close but does not exactly match the initial account value of 400. Checking further might show divergences.

- [tex]\( f(x) = 401.19 \cdot (1.08)^x \)[/tex]
- [tex]\( f(0) = 401.19 \cdot (1.08)^0 = 401.19 \)[/tex]
- This closely matches the initial account value of 400. Checking more values:
- [tex]\( f(1) = 401.19 \cdot (1.08)^1 = 401.19 \cdot 1.08 \approx 433.29 \)[/tex]
- [tex]\( f(2) = 401.19 \cdot (1.08)^2 \approx 467.95 \)[/tex]
- [tex]\( f(3) = 401.19 \cdot (1.08)^3 \approx 504.39 \)[/tex]
- [tex]\( f(4) = 401.19 \cdot (1.08)^4 \approx 543.86 \)[/tex]
- [tex]\( f(5) = 401.19 \cdot (1.08)^5 \approx 587.37 \)[/tex]

These values closely match the provided account values when rounded to the nearest hundredth and are closer than any other function candidate. Thus, the exponential function that models the data is:

[tex]\[ f(x) = 401.19 \cdot (1.08)^x \][/tex]
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