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Which is the setup to find the temperature 10 hours after sundown using the trendline [tex]y=-1.6x+72.4[/tex]?

A.
[tex]\[
\begin{array}{l}
(10)=-1.6x+72.4 \\
-72.4+10=-1.6x+72.4-72.4 \\
-62.4=-1.6x \\
\frac{-62.4}{1.6}=\frac{-1.6}{-1.6}x \\
-39=x
\end{array}
\][/tex]

B.
[tex]\[
\begin{array}{l}
y=-1.6(10)+72.4 \\
y=-16+72.4 \\
y=56.4
\end{array}
\][/tex]

C.
[tex]\[
\begin{array}{l}
(10)=-1.6x+72.4 \\
-72.4+10=-1.6x+72.4-72.4 \\
-62.4=-1.6x \\
\frac{-62.4}{1}=\frac{-1.6}{4}x
\end{array}
\][/tex]



Answer :

GinnyAnswer
To find the temperature 10 hours after sundown using the trendline [tex]\( y = -1.6x + 72.4 \)[/tex], follow these steps:

1. Identify the value of [tex]\( x \)[/tex] given in the problem:
[tex]\[ x = 10 \text{ hours after sundown} \][/tex]

2. Substitute [tex]\( x = 10 \)[/tex] into the trendline equation [tex]\( y = -1.6x + 72.4 \)[/tex]:
[tex]\[ y = -1.6(10) + 72.4 \][/tex]

3. Perform the multiplication:
[tex]\[ y = -16 + 72.4 \][/tex]

4. Add the results to find the temperature:
[tex]\[ y = 56.4 \][/tex]

Hence, the temperature 10 hours after sundown is [tex]\( 56.4 \)[/tex] degrees.

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