Answer:
Step-by-step explanation:
If slope of two lines are equal, then they are parallel lines.
If product of the slope of two lines is (-1), then they are perpendicular lines.
C) y = x + 2
Slope m1 = 1
y = -x + 3
Slope m2 = -1
m1 * m2 = 1 * (-1) = -1
These are perpendicular lines.
D) y = 3x + 2
Slope = m1 = 3
y = 3x - 2
Slope = m2 = 3
Slopes are equal. So, they are parallel lines.
E)y= 3
This line is parallel to x -axis.
x = 4
This line is parallel to y-axis.
So, both lines are perpendicular to each other.
F) y = x + 8
slope m1 = 1
y = -x + 3
Slope = m2 = -1
M1 * m2 = 1 * (-1) = -1
So, These are perpendicular lines
Subtract the cost of dog food with your total amount and then find the percentage of that amount based on the total you have so in this case it would be 85 / 1695 * 100 and then you will get your percentage
Answer:
Step-by-step explanation:
28 students all give $128
so,
![\sqrt[28]{128}](https://tex.z-dn.net/?f=%5Csqrt%5B28%5D%7B128%7D)
= v v=6
Each student gave 6 dollars. This is true due to the fact that
=6
28X6=128
C is the correct answer hope this helps you and have a great day
Answer:
a) y = 0.74x + 18.99; b) 80; c) r = 0.92, r² = 0.85; r² tells us that 85% of the variance in the dependent variable, the final average, is predictable from the independent variable, the first test score.
Step-by-step explanation:
For part a,
We first plot the data using a graphing calculator. We then run a linear regression on the data.
In the form y = ax + b, we get an a value that rounds to 0.74 and a b value that rounds to 18.99. This gives us the equation
y = 0.74x + 18.99.
For part b,
To find the final average of a student who made an 83 on the first test, we substitute 83 in place of x in our regression equation:
y = 0.74(83) + 18.99
y = 61.42 + 18.99 = 80.41
Rounded to the nearest percent, this is 80.
For part c,
The value of r is 0.92. This tells us that the line is a 92% fit for the data.
The value of r² is 0.85. This is the coefficient of determination; it tells us how much of the dependent variable can be predicted from the independent variable.
