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:
![\begin{bmatrix}\mathrm{Solution:}\:&\:x\le \frac{1200}{499}\:\\ \:\mathrm{Decimal:}&\:x\le \:2.40480\dots \\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:\frac{1200}{499}]\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Ax%5Cle%20%5Cfrac%7B1200%7D%7B499%7D%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BDecimal%3A%7D%26%5C%3Ax%5Cle%20%5C%3A2.40480%5Cdots%20%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%28-%5Cinfty%20%5C%3A%2C%5C%3A%5Cfrac%7B1200%7D%7B499%7D%5D%5Cend%7Bbmatrix%7D)
Step-by-step explanation:

A) yes
B) yes
C) no
For each of these, substitute the value of x in the ordered pair into x in the function.
For A, x = -5; -5<2, so the piece of the function we want is f(x) = 3. In our ordered pair, y=f(x)=3, so yes, it is a solution.
For B, x = 2; 2≤2<6, so the piece of the function we want is f(x) = -x+1. In our ordered pair, y=f(x)=-1; -2+1=-1, so yes, it is a solution.
For C, x = 8; 8≥6, so the piece of the function we want is f(x) = x. In our ordered pair, y=f(x)=-7; -7≠8, so no, it is not a solution.
Answer:
Option A.
; grows approximately at a rate of 0.4% daily
Step-by-step explanation:
we have

where
f(x) the number of weeds in the garden
x ----> the number of weeks
Calculate how quickly the weeds grow each day
Remember that a week is equal to seven days
so

Using the law of exponents
b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x
so
![f(x)=86[(1.29)^{\frac{1}{7}}]^{x}](https://tex.z-dn.net/?f=f%28x%29%3D86%5B%281.29%29%5E%7B%5Cfrac%7B1%7D%7B7%7D%7D%5D%5E%7Bx%7D)
![f(x)=86[1.04]^{x}](https://tex.z-dn.net/?f=f%28x%29%3D86%5B1.04%5D%5E%7Bx%7D)
therefore
The rate is approximately
1.04=1+r
r=1.04-1=0.04=4% daily