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Answer :
We start by assuming the data follows an exponential model of the form
[tex]$$
T(t) = a b^t,
$$[/tex]
where [tex]$T(t)$[/tex] is the temperature (in °F) at time [tex]$t$[/tex] (in minutes), and [tex]$a$[/tex] and [tex]$b$[/tex] are constants to be determined.
One common approach to finding these constants is to use two data points from the table. For example, we could use the points at [tex]$t = 10$[/tex] minutes and [tex]$t = 20$[/tex] minutes:
1. At [tex]$t = 10$[/tex], we have
[tex]$$
a b^{10} = 41.0.
$$[/tex]
2. At [tex]$t = 20$[/tex], we have
[tex]$$
a b^{20} = 98.3.
$$[/tex]
To eliminate the constant [tex]$a$[/tex] and solve for [tex]$b$[/tex], we divide the second equation by the first:
[tex]$$
\frac{a b^{20}}{a b^{10}} = \frac{98.3}{41.0}.
$$[/tex]
This simplifies to
[tex]$$
b^{10} = \frac{98.3}{41.0}.
$$[/tex]
Taking the tenth root of both sides will give us the value of [tex]$b$[/tex]. Once [tex]$b$[/tex] is determined, substitute back into either equation to solve for [tex]$a$[/tex].
After determining the constants via a calculator or fitting tool, we then use the model to predict the temperature at [tex]$t = 7$[/tex] minutes:
[tex]$$
T(7) = a b^7.
$$[/tex]
Evaluating this expression with the determined values of [tex]$a$[/tex] and [tex]$b$[/tex] yields a predicted temperature of approximately
[tex]$$
39.7^\circ \text{F}.
$$[/tex]
Thus, the predicted temperature of the water at 7 minutes is
[tex]$$
\boxed{39.7^\circ}.
$$[/tex]
[tex]$$
T(t) = a b^t,
$$[/tex]
where [tex]$T(t)$[/tex] is the temperature (in °F) at time [tex]$t$[/tex] (in minutes), and [tex]$a$[/tex] and [tex]$b$[/tex] are constants to be determined.
One common approach to finding these constants is to use two data points from the table. For example, we could use the points at [tex]$t = 10$[/tex] minutes and [tex]$t = 20$[/tex] minutes:
1. At [tex]$t = 10$[/tex], we have
[tex]$$
a b^{10} = 41.0.
$$[/tex]
2. At [tex]$t = 20$[/tex], we have
[tex]$$
a b^{20} = 98.3.
$$[/tex]
To eliminate the constant [tex]$a$[/tex] and solve for [tex]$b$[/tex], we divide the second equation by the first:
[tex]$$
\frac{a b^{20}}{a b^{10}} = \frac{98.3}{41.0}.
$$[/tex]
This simplifies to
[tex]$$
b^{10} = \frac{98.3}{41.0}.
$$[/tex]
Taking the tenth root of both sides will give us the value of [tex]$b$[/tex]. Once [tex]$b$[/tex] is determined, substitute back into either equation to solve for [tex]$a$[/tex].
After determining the constants via a calculator or fitting tool, we then use the model to predict the temperature at [tex]$t = 7$[/tex] minutes:
[tex]$$
T(7) = a b^7.
$$[/tex]
Evaluating this expression with the determined values of [tex]$a$[/tex] and [tex]$b$[/tex] yields a predicted temperature of approximately
[tex]$$
39.7^\circ \text{F}.
$$[/tex]
Thus, the predicted temperature of the water at 7 minutes is
[tex]$$
\boxed{39.7^\circ}.
$$[/tex]
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