3/4 is 0.75.
3/4 is larger than 0.7
Y-3=(1/2)(x+6). just solve for y
y-3=1/2x +3
y = 1/2x + 6
To find out if the points lie on the line y = 1/3x + 9, plug in the x value of the coordinate into the equation and see if the y value results in the same number
A. (0,9)
y = 1/3x + 9 Plug in 0 for "x"
y = 1/3(0) + 9
y = 9 This point is on the line
B. (-9,6)
y = 1/3x + 9 Plug in -9 for "x"
y = 1/3(-9) + 9
y = -9/3 + 9
y = -3 + 9
y = 6 This point is on the line
C. (3,10)
y = 1/3x + 9 Plug in 3 for "x"
y = 1/3(3) + 9
y = 3/3 + 9
y = 1 + 9
y = 10 This point is on the line
D. (12, -5)
y = 1/3x + 9 Plug in 12 for "x"
y = 1/3(12) + 9
y = 12/3 + 9
y = 4 + 9
y = 13 This point is not on the line
Your answer is A, B, and C
One day, you go to a store and buy a pencil. It costs $0.25.
You notice the unit price $0.25/pencil.
The next day you buy 3 pencils, they cost $0.75.
You notice the unit price is $0.75/(3 pencils) = $0.25/pencil.
For each extra pencil you buy, the increase in price is always $0.25.
This suggests a linear relation between the number of pencils, p, and the cost of the pencils, c.
You use a linear equation to give you the cost of pencils based on the number of pencils.
c = 0.25p
One day, you happen to have $4.50 in your pocket.
You'd like to buy 15 pencils, but you are not sure you have enough money.
You use your handy linear equation, and you replace p, the number of pencils, with 15 to find their cost, c.
c = 0.25p
c = 0.25 * 15
c = 3.75
After using your equation, you see that 15 pencils cost $3.75, and since you have $4.50 you have enough money to buy them. You're very happy about this, and, as soon as you get home, you write this experience in your journal, using one of the new pencils you just bought, and you make it a point to include the linear equation that helped you.
Answer:
Is not appropiate to refer a estimation or a statistic as a paramter because the statistic just give informaation about the sample selected and not about all the population of interest. What we can do is inference with this sample proportion or confidence intervals in order to see on what limits our real parameter of interest p lies.
Step-by-step explanation:
Description in words of the parameter p
represent the real population proportion of students who went Home for winter break
represent the estimated proportion of students who went Home for winter break
n is the sample size selected
The population proportion have the following distribution
Solution to the problem
For this case we assume that the proportion given 0.35 is an estimation for the real parameter of interest p, that means 
On this case the estimated proportion is calculated from the following formula:
Where X are the people in the sam with the characteristic desired (students who went Home for winter break) and n the sample size selected.
Is not appropiate to refer a estimation or a statistic as a paramter because the statistic just give informaation about the sample selected and not about all the population of interest. What we can do is inference with this sample proportion or confidence intervals in order to see on what limits our real parameter of interest p lies.
