Graph f(x) = x and g(x) = and g(x) =-1/3x+2 x on the same coordinate plane.
1 answer:
Hey there!
Let's think of both of these functions as two different slope-intercept equations (y=mx+b). Don't let all of the fs and gs confuse you; those are just showing that they are two different functions!
Our function f has an equation of y=x, which is a diagonal line that passes through the origin and goes across each small grid on the graph diagonally.
Our function g has an equation of y=-1/3x+2, which means that it will intersect with the y-axis at two units above the origin and for every three whole steps it goes to the right, it will have gradually gone down one whole step as well.
Attached to my answer is what a graph of it would kind of look like. Sorry if the lines don't look too much like a line it's a bit difficult to draw with a mouse haha!
I hope that this helps! Have a wonderful day!
You might be interested in
Given:
In parallelogram ABCD, two of its vertices are A(-4,0) and B(0,3).
To find:
The equation that represents a line that contain CD.
Solution:
We have,
A(-4,0) and B(0,3)
Slope of AB is
The slope of line AB is .
Opposite sides of a parallelogram are parallel and slopes of parallel lines are equal.
In parallelogram ABCD, AB and CD are opposite sides. So, their slopes must be equal.
Slope of line AB = Slope of line CD =
The slope intercept form of a line is
Where, m is slope and b is y-intercept.
Slope of line CD is , it means the line must be of the form
Coefficient of x is only in option a.
Therefore, the correct option is a.
Answer:
Step-by-step explanation:
Hello!
The variable of interest is X: chest circumference of a Scottish man.
X≈N(μ;δ²)
μ= 40 inches
δ= 2 inches
The empirical rule states that
68% of the distribution lies within one standard deviation of the mean: μ±δ= 0.68
95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95
99% of the distribution lies within 3 standard deviations of the mean: μ±3δ= 0.99
a)
The 58% that falls closest to the mean can also be referred to as the middle 58% of the distribution, assuming that both values are equally distant from the mean.
P(a≤X≤b)= 0.58
If 1-α= 0.58, then the remaining proportion α= 0.42 is divided in two equal tails α/2= 0.21.
The accumulated proportion until "a" is 0.21 and the accumulated proportion until "b" is 0.21 + 0.58= 0.79 (See attachment)
P(X≤a)= 0.21
P(X≤b)= 0.79
Using the standard normal distribution, you can find the corresponding values for the accumulated probabilities, then using the information of the original distribution:
P(Z≤zᵃ)= 0.21
zᵃ= -0.806
P(Z≤zᵇ)= 0.79
zᵇ= 0.806
Using the standard normal distribution Z= (X-μ)/δ you "transform" the values of Z to values of chest circumference (X):
zᵃ= (a-μ)/δ
zᵃ*δ= a-μ
a= (zᵃ*δ)+μ
a= (-0.806*2)+40= 38.388
and
zᵇ= (b-μ)/δ
zᵇ*δ= b-μ
b= (zᵇ*δ)+μ
b= (0.806*2)+40= 41.612
58% of the chest measurements will be within 38.388 and 41.612 inches.
b)
The measurements of the 2.5% men with the smallest chest measurements, can also be interpreted as the "bottom" 2.5% of the distribution, the value that separates the bottom 2.5% of the distribution from the 97.5%, symbolically:
P(X≤b)= 0.025 (See attachment)
Now you have to look under the standard normal distribution the value of z that accumulates 0.025 of the distribution:
P(Z≤zᵇ)= 0.025
zᵇ= -1.960
Now you reverse the standardization to find the value of chest circumference:
zᵇ= (b-μ)/δ
zᵇ*δ= b-μ
b= (zᵇ*δ)+μ
b= (-1.960*2)+40= 36.08
The chest measurement of the 2.5% smallest chest measurements is 36.08 inches.
c)
Using the empirical rule:
95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95
(μ-2δ) ≤ Xc ≤ (μ+2δ)=0.95 ⇒ (40-4) ≤ Xc ≤ (40+4)= 0.95 ⇒ 36 ≤ Xc ≤ 44= 0.95
d)
The measurements of the 16% of the men with the largest chests in the population or the "top" 16% of the distribution:
P(X≥d)= 0.16
P(X≤d)= 1 - 0.16
P(X≤d)= 0.84
First, you look for the value that accumulates 0.84 of probability under the standard normal distribution:
P(Z≤zd)= 0.84
zd= 0.994
Now you reverse the standardization to find the value of chest circumference:
zd= (d-μ)/δ
zd*δ= d-μ
d= (zd*δ)+μ
d= (0.994*2)+40= 41.988
The measurements of the 16% of the men with larges chess are at least 41.988 inches.
I hope this helps!
