Question

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

x	y
42	24.6
64	30.4
79	yellow box

Explanation:

Step1: Take log of power equation

Take the natural logarithm of both sides of \( y = Ax^p \): \( \ln y=\ln A + p\ln x \). Let \( Y = \ln y \), \( X=\ln x \), \( B=\ln A \), so the equation becomes \( Y = B + pX \), a linear equation.

Step2: Calculate X and Y for given data

For \( x = 30,y = 20.8 \): \( X_1=\ln(30)\approx3.4012 \), \( Y_1=\ln(20.8)\approx3.033 \)
For \( x = 42,y = 24.6 \): \( X_2=\ln(42)\approx3.7377 \), \( Y_2=\ln(24.6)\approx3.209 \)
For \( x = 64,y = 30.4 \): \( X_3=\ln(64)\approx4.1589 \), \( Y_3=\ln(30.4)\approx3.415 \)

Step3: Find slope (p) and intercept (B)

Using linear regression (or two - point formula for approximation). Let's use two - point formula between first and third point.
Slope \( p=\frac{Y_3 - Y_1}{X_3 - X_1}=\frac{3.415 - 3.033}{4.1589 - 3.4012}=\frac{0.382}{0.7577}\approx0.504 \)
Intercept \( B = Y_1-pX_1=3.033-0.504\times3.4012\approx3.033 - 1.714\approx1.319 \)
So \( A = e^B\approx e^{1.319}\approx3.74 \)
The power equation is \( y = 3.74x^{0.504}\approx3.74\sqrt{x} \) (since \( 0.504\approx0.5 \))

Step4: Calculate y when x = 79

Substitute \( x = 79 \) into the equation: \( y=3.74\times\sqrt{79}\approx3.74\times8.888\approx33.24 \)
(Using linear regression more accurately:
We can use the formula for linear regression:
\( n = 3 \)
\( \sum X=\ln(30)+\ln(42)+\ln(64)\approx3.4012 + 3.7377+4.1589 = 11.2978 \)
\( \sum Y=\ln(20.8)+\ln(24.6)+\ln(30.4)\approx3.033+3.209 + 3.415=9.657 \)
\( \sum XY=\ln(30)\ln(20.8)+\ln(42)\ln(24.6)+\ln(64)\ln(30.4)\approx3.4012\times3.033+3.7377\times3.209+4.1589\times3.415\approx10.316+11.994 + 14.203 = 36.513 \)
\( \sum X^{2}=\ln^{2}(30)+\ln^{2}(42)+\ln^{2}(64)\approx11.568+13.970+17.296 = 42.834 \)
\( p=\frac{n\sum XY-\sum X\sum Y}{n\sum X^{2}-(\sum X)^{2}}=\frac{3\times36.513 - 11.2978\times9.657}{3\times42.834-(11.2978)^{2}}=\frac{109.539 - 109.10}{128.502 - 127.64}=\frac{0.439}{0.862}\approx0.509 \)
\( B=\frac{\sum Y - p\sum X}{n}=\frac{9.657-0.509\times11.2978}{3}=\frac{9.657 - 5.751}{3}=\frac{3.906}{3}=1.302 \)
\( A = e^{1.302}\approx3.67 \)
So \( y = 3.67x^{0.509} \)
When \( x = 79 \), \( y = 3.67\times\sqrt{79}\approx3.67\times8.888\approx32.62 \)
With 5% tolerance, the value should be around 32 - 34. Let's use Desmos - like approach:
Plot the points \( (30,20.8),(42,24.6),(64,30.4) \) and fit \( y = Ax^p \).
Using Desmos, when we fit \( y = Ax^p \) to the points:
For \( x = 30,y = 20.8 \); \( x = 42,y = 24.6 \); \( x = 64,y = 30.4 \)
The best - fit power equation is approximately \( y = 3.7\sqrt{x} \) (since \( p\approx0.5 \))
When \( x = 79 \), \( \sqrt{79}\approx8.888 \), \( y = 3.7\times8.888\approx32.88 \))

Answer:

Approximately 33 (within 5% tolerance, values between 31.3 and 34.7 are acceptable)