ŷ= 1.795x +2.195 is the equation for the line of best fit for the data
<h3>How to use regression to find the equation for the line of best fit?</h3>
Consider the table in the image attached:
∑x = 29, ∑y = 74, ∑x²= 125, ∑xy = 288, n = 10 (number data points)
The linear regression equation is of the form:
ŷ = ax + b
where a and b are the slope and y-intercept respectively
a = ( n∑xy -(∑x)(∑y) ) / ( n∑x² - (∑x)² )
a = (10×288 - 29×74) / ( 10×125-29² )
= 2880-2146 / 1250-841
= 734/409
= 1.795
x' = ∑x/n
x' = 29/10 = 2.9
y' = ∑y/n
y' = 74/10 = 7.4
b = y' - ax'
b = 7.4 - 1.795×2.9
= 7.4 - 5.2055
= 2.195
ŷ = ax + b
ŷ= 1.795x +2.195
Therefore, the equation for the line of best fit for the data is ŷ= 1.795x +2.195
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Answer:
typing this answer to get points hope u get ur answer soon :)
Step-by-step explanation:
Answer:
B. 8x^2-6x-7
Step-by-step explanation:
All you have to do is combine the like terms.
Like terms are the terms that have the same variable and same exponent.
The like terms in this equation are 6x^2 and 2x^2, 11x and -17x, and -3 and -4.
When you add 6x^2 and 2x^2, you get 8x^2
When you add 11x and -17x, you get -6x
When you add -3 and -4, you get -7.
Putting these all in order, your answer is
8x^2 - 6x - 7
Answer:15 percent
Step-by-step explanation:
(.75) / 5 = 3/20 = 15/100 = 15%
Answer:
$385,628
Depreciation = ($2,000,000 - $250,000)/10 = $175,000
$175,000 x .39 = $68250 tax savings each period.
Across the entire project, these savings will constitute a 10 period annuity.
Pmt = 68,250, FV = 0, I = 12, N = 10, PV = 385,627.72
Step-by-step explanation:
