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erica [24]
4 years ago
12

A study took a random sample often married couples, and asked the husband's and wife's age. The scatterplot shows the associatio

n between the ages of the husbands and wives. Not surprisingly, older husbands tend to have older wives, so the data has a positive linear association. A researcher finds that the least squares regression equation for this relationship is ŷ= 0.89 x + 2.62.
Suppose you interview a married man who has lived to the ripe old age of 100, and still married to his wife. using the least squares regression equation, predict his wife's age.

A. 100
B. 89
C. 91.62
D. 86.38​

—my answer is option (c.) 91.62
Any reviews would be much appreciated...

Mathematics
2 answers:
gulaghasi [49]4 years ago
7 0
I agree I personally think it’s C too! :)
lidiya [134]4 years ago
7 0
I agree :)! The correct answer is C, 91.62. Well.. I think. lol.
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1. Divide wire b in parts x and b-x. 

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= \frac{1}{2}sin60^{o}side*side=   \frac{1}{2} \frac{ \sqrt{3} }{2}  (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}
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4. Let f(x)=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)=\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b

x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

6. So one part is \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) } and the other part is b-\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

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In this problem, we have that:

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133 = 189 - 2*28

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