Answer:
y = 14x
Step-by-step explanation:
Use the direct variation equation, y = kx
Plug in 7 as y and 0.5 as x, and solve for k:
y = kx
7 = k(0.5)
14 = k
Plug this into the equation:
y = kx
y = 14x
So, the equation of variation is y = 14x
Answer:
It is not normally distributed as it has it main concentration in only one side.
Step-by-step explanation:
So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).
So, let us begin the groupings into their different classes, shall we?
Data given:
0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.
(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.
(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31
(3). 0.4 - 0.6 : there are 2 values that falls into this category.
(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.
Class interval frequency.
0.00 - 0.20. 15.
0.20 - 0.40. 2.
0.4 - 0.6. 2.
D
1/8 = 0.125
0.125*5=0.625
0.625+1=1 5/8
If you were to plot the function y=-x+6, both functions meet at the point (4,2)
A
Step-by-step explanation:First, subtract
2
π
r
2
from each side of the equation to isolate the
h
term:
S
−
2
π
r
2
=
2
π
r
h
+
2
π
r
2
−
2
π
r
2
S
−
2
π
r
2
=
2
π
r
h
+
0
S
−
2
π
r
2
=
2
π
r
h
Now, divide each side of the equation by
2
π
r
to solve for
h
:
S
−
2
π
r
2
2
π
r
=
2
π
r
h
2
π
r
S
−
2
π
r
2
2
π
r
=
2
π
r
h
2
π
r
S
−
2
π
r
2
2
π
r
=
h
h
=
S
−
2
π
r
2
2
π
r
Or
h
=
S
2
π
r
−
2
π
r
2
2
π
r
h
=
S
2
π
r
−
2
π
r
2
2
π
r
h
=
S
2
π
r
−
r
2
r
h
=
S
2
π
r
−
r