which of the following equations best models the information in the table?
$y = 0.006x^{2}+0.245x + 0.012$
$y = 0.715x^{2}+5.982x + 1.591$

Question

Accepted Answer

# Explanation:
## Step1: Substitute $x = 20$ into the first equation
For $y = 0.006x^{2}+0.245x + 0.012$, when $x = 20$:
$$
\begin{align*}
y&=0.006\times(20)^{2}+0.245\times20+0.012\
&=0.006\times400 + 4.9+0.012\
&=2.4+4.9+0.012\
&=7.312
\end{align*}
$$
## Step2: Substitute $x = 20$ into the second equation
For $y = 0.715x^{2}+5.982x + 1.591$, when $x = 20$:
$$
\begin{align*}
y&=0.715\times(20)^{2}+5.982\times20+1.591\
&=0.715\times400+119.64 + 1.591\
&=286+119.64+1.591\
&=407.231
\end{align*}
$$
## Step3: Compare with the table value
The table value for $x = 20$ is $y = 7.5$. The result from $y = 0.006x^{2}+0.245x + 0.012$ ($y = 7.312$) is much closer to $7.5$ than the result from $y = 0.715x^{2}+5.982x + 1.591$ ($y = 407.231$).

# Answer:
$y = 0.006x^{2}+0.245x + 0.012$

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QUESTION IMAGE

Question

Explanation:

Step1: Substitute \(x = 20\) into the first equation

Step2: Substitute \(x = 20\) into the second equation

Step3: Compare with the table value

Answer: