Question

extrapolate!
we recommend using the desmos graphing calculator.
answers have 5% tolerance.
assume data fits a power equation: ( y = ax^b ).

x	y
32	83.7
41	107
52	■

Explanation:

Step1: Take log of power equation

Take natural logarithm of \( y = Ax^b \), we get \( \ln y=\ln A + b\ln x \). Let \( Y = \ln y \), \( X=\ln x \), \( C=\ln A \), so \( Y = C + bX \), a linear equation.

Step2: Calculate X and Y for given data

For \( x = 20,y = 52.3 \): \( X_1=\ln(20)\approx2.9957 \), \( Y_1=\ln(52.3)\approx3.957 \)
For \( x = 32,y = 83.7 \): \( X_2=\ln(32)\approx3.4657 \), \( Y_2=\ln(83.7)\approx4.429 \)
For \( x = 41,y = 107 \): \( X_3=\ln(41)\approx3.7136 \), \( Y_3=\ln(107)\approx4.673 \)

Step3: Find slope \( b \) and intercept \( C \)

Using linear regression (or two - point formula, here we use three points for better accuracy). The slope \( b=\frac{\sum(X_i - \bar{X})(Y_i - \bar{Y})}{\sum(X_i - \bar{X})^2} \)
First, calculate \( \bar{X}=\frac{2.9957 + 3.4657+3.7136}{3}=\frac{10.175}{3}\approx3.3917 \)
\( \bar{Y}=\frac{3.957 + 4.429+4.673}{3}=\frac{13.059}{3}=4.353 \)
\( \sum(X_i - \bar{X})(Y_i - \bar{Y})=(2.9957 - 3.3917)(3.957 - 4.353)+(3.4657 - 3.3917)(4.429 - 4.353)+(3.7136 - 3.3917)(4.673 - 4.353) \)
\(=(- 0.396)(-0.396)+(0.074)(0.076)+(0.3219)(0.32) \)
\(=0.1568 + 0.0056+0.1030 = 0.2654 \)
\( \sum(X_i - \bar{X})^2=(-0.396)^2+(0.074)^2+(0.3219)^2=0.1568 + 0.0055+0.1036 = 0.2659 \)
\( b=\frac{0.2654}{0.2659}\approx1 \)
Then, using \( Y = C + bX \), take the first point: \( 3.957=C + 1\times2.9957\Rightarrow C=3.957 - 2.9957 = 0.9613 \)
Since \( C=\ln A \), then \( A = e^{0.9613}\approx2.614 \)
So the power equation is \( y = 2.614x^{1} \) (approximately, since \( b\approx1 \))

Step4: Calculate y when x = 52

Substitute \( x = 52 \) into \( y = 2.614x \), we get \( y=2.614\times52 = 135.928 \)
(We can also use more accurate linear regression. Using a calculator for linear regression of \( (\ln x,\ln y) \):
The data points for \( (\ln x,\ln y) \) are \( (2.9957,3.957),(3.4657,4.429),(3.7136,4.673) \)
Using linear regression formula \( b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^{2}-(\sum X)^{2}} \), \( n = 3 \)
\( \sum X=2.9957 + 3.4657+3.7136 = 10.175 \)
\( \sum Y=3.957 + 4.429+4.673 = 13.059 \)
\( \sum XY=2.9957\times3.957+3.4657\times4.429+3.7136\times4.673 \)
\(=11.85 + 15.35+17.35 = 44.55 \)
\( \sum X^{2}=2.9957^{2}+3.4657^{2}+3.7136^{2}=8.97 + 12.01+13.79 = 34.77 \)
\( b=\frac{3\times44.55-10.175\times13.059}{3\times34.77-(10.175)^{2}}=\frac{133.65 - 132.87}{104.31 - 103.53}=\frac{0.78}{0.78}=1 \)
\( C=\frac{\sum Y - b\sum X}{n}=\frac{13.059-1\times10.175}{3}=\frac{2.884}{3}\approx0.961 \), \( A = e^{0.961}\approx2.61 \)
So \( y = 2.61x \), when \( x = 52 \), \( y = 2.61\times52=135.72 \))

Answer:

\( 136 \) (with 5% tolerance, the range is \( 136\times(1 - 0.05)=129.2 \) to \( 136\times(1 + 0.05)=142.8 \), and our calculation is within this range)