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Answer :
We are given the quadratic regression equation
[tex]$$
y = 0.008x^2 + 0.518x + 131.886.
$$[/tex]
We need to predict the [tex]$y$[/tex]-value when [tex]$x = 200$[/tex]. Follow these steps:
1. Begin by substituting [tex]$x = 200$[/tex] into the equation:
[tex]$$
y = 0.008(200)^2 + 0.518(200) + 131.886.
$$[/tex]
2. Calculate the quadratic term:
[tex]$$
0.008(200)^2 = 0.008 \times 40000 = 320.
$$[/tex]
3. Calculate the linear term:
[tex]$$
0.518(200) = 103.6.
$$[/tex]
4. The constant term remains as it is:
[tex]$$
131.886.
$$[/tex]
5. Finally, add all these results together:
[tex]$$
y = 320 + 103.6 + 131.886 = 555.486.
$$[/tex]
Rounding to one decimal place, we have [tex]$y \approx 555.5$[/tex].
Thus, the correct answer is [tex]$\boxed{555.5}$[/tex], which corresponds to option d.
[tex]$$
y = 0.008x^2 + 0.518x + 131.886.
$$[/tex]
We need to predict the [tex]$y$[/tex]-value when [tex]$x = 200$[/tex]. Follow these steps:
1. Begin by substituting [tex]$x = 200$[/tex] into the equation:
[tex]$$
y = 0.008(200)^2 + 0.518(200) + 131.886.
$$[/tex]
2. Calculate the quadratic term:
[tex]$$
0.008(200)^2 = 0.008 \times 40000 = 320.
$$[/tex]
3. Calculate the linear term:
[tex]$$
0.518(200) = 103.6.
$$[/tex]
4. The constant term remains as it is:
[tex]$$
131.886.
$$[/tex]
5. Finally, add all these results together:
[tex]$$
y = 320 + 103.6 + 131.886 = 555.486.
$$[/tex]
Rounding to one decimal place, we have [tex]$y \approx 555.5$[/tex].
Thus, the correct answer is [tex]$\boxed{555.5}$[/tex], which corresponds to option d.
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