Answer:
a. or 
Step-by-step explanation:
Slope is equal to y1-y2 over x1-x2.
So the equation would be:
2-(-1)=2+1=3
-3-(-1)=-3+1=-2
So the answer is:
30?
30pi = 2pi r
30pi / 2pi = 15
15 = r
double the radius to get the diameter
15)2 =30
Answer: <em>x=4</em>
Step-by-step explanation:
<em>1+3x=x+9</em>
<em> -x -x</em>
<em>1+2x=9</em>
<em>-1 -1</em>
<em>2x=8</em>
<em>Divide by 2 on each side</em>
<em>Final result</em>
<em>x=4</em>
Answer:
45
Step-by-step explanation:
AEF is a similar triangle to ABC. that means it has the same angles, and the sides (and all other lines in the triangle) are scaled from the ABC length to the AEF length by the same factor f.
now, what is f ?
we know this from the relation of AC to FA.
FA = 12 mm
AC = 12 + 28 = 40 mm
so, going from AC to FA we multiply AC by f so that
AC × f = FA
40 × f = 12
f = 12/40 = 3/10
all other sides, heights, ... if ABC translate to their smaller counterparts in AEF by that multiplication with f (= 3/10).
the area of a triangle is
baseline × height / 2
aABC = 500
and because of the similarity we don't need to calculate the side and height in absolute numbers. we can use the relative sizes by referring to the original dimensions and the scaling factor f.
baseline small = baseline large × f
height small = height large × f
we know that
baseline large × height large / 2 = 500
baseline large × height large = 1000
aAEF = baseline small × height small / 2 =
= baseline large × f × height large × f / 2 =
= baseline large × height large × f² / 2 =
= 1000 × f² / 2 = 500 × f² = 500 ×(3/10)² =
= 500 × 9/100 = 5 × 9 = 45 mm²
Answer:
Step-by-step explanation:
a)
Confidence interval in less than symbol expressed as
Where 
is sample mean and 
is margin of error.
b)
The given t interval is 
That is 
and 
Solve these two equation by adding together.
Solve this value of \bar{x} in equation and solve for
Best point estimate of 
Best point estimate of margin of error = 0.193
c)
Since sample size = 100 which is sufficiently large (Greater than 30) , it is no need to confirm that
sample data appear to be form a population with normal distribution.
