A cone always contain a right angle.
To solve for the volume of a cone, we need to get the radius of the cone and its height.
Volume of a cone = π r² h/3
radius is the line from the center of the circle of the cone going to the edge of the circle.
height is the measurement of the cone from its pointy tip going downward.
Assuming the radius of the cone is 3 in. and its height is 6 inches.
Volume of a cone = π * r² * h/3
V = 3.14 * (3in)² * 6in/3
V = 3.14 * 9in² * 2in
V = 56.52 in³
Answer:
sigma should be used
Step-by-step explanation:
Given that The mean length of time in jail from the survey was four years with a standard deviation of 1.9 years.
The above given is for sample of 27 size.
For hypothesis test to compare mean of sample with population we can use either population std dev or sample std dev.
But once population std deviation is given, we use only that as that would be more reliable.
So here we can use population std deviation 1.4 only.
If population std deviation is used we can use normality and do Z test
Answer:
The answer to your question is the first graph.
Step-by-step explanation:
It seems that the first graph is correct because the height of the bars is equal to the grade get by the students.
The second graph is incorrect because the height of the bars is different from the grade the students got. Only the bar of Levi is correct.
The third graph is incorrect, just observe the height of John's bar to determine that the graph is wrong.
Also, the last graph is incorrect, in this graph, all the bars are different from the grade the students got.
Answer:
Step-by-step explanation:
21.8% of x = 51,800
0.218x =51,800
x= 51,800/0.218 = 237,615
Answer:
A - Rectangle B - Square
C - Parallelogram D - Rhombus
Explanation:
We are given
A
(
1
,
2
)
,
B
(
2
,
−
2
)
and hence
A
B
=
√
(
2
−
1
)
2
+
(
−
2
−
2
)
2
=
√
17
. Further slope of
A
B
is
−
2
−
2
2
−
1
=
−
4
1
=
−
4
.
Case A -
C
(
−
6
,
−
4
)
,
D
(
−
7
,
0
)
As
C
D
=
√
(
−
7
−
(
−
6
)
)
2
+
(
0
−
(
−
4
)
)
2
=
√
17
and slope of
C
D
is
0
−
(
−
4
)
−
7
−
(
−
6
)
=
4
−
1
=
−
4
As
A
B
=
C
D
and
A
B
||
C
D
slopes being equal, ABCD is a parallelogram.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+6)^2+(y+4)^2-0.08)((x+7)^2+y^2-0.08)=0 [-10, 10, -5, 5]}
Case B -
C
(
6
,
−
1
)
,
D
(
5
,
3
)
As
C
D
=
√
(
5
−
6
)
2
+
(
3
−
(
−
1
)
)
2
=
√
17
and slope of
C
D
is
0
−
(
−
4
)
−
7
−
(
−
6
)
=
4
−
1
=
−
4
Further,
B
C
=
√
(
6
−
2
)
2
+
(
−
1
−
(
−
2
)
)
2
=
√
17
and slope of
B
C
is
−
1
−
(
−
2
)
6
−
2
=
1
4
As
B
C
=
A
B
and they are perpendicular (as product of slopes is
−
1
), ABCD is a square.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-6)^2+(y+1)^2-0.08)((x-5)^2+(y-3)^2-0.08)=0 [-10, 10, -5, 5]}
Case C -
C
(
−
1
,
−
4
)
,
D
(
−
2
,
0
)
As mid point of
A
C
is
(
1
−
1
2
,
2
−
4
2
)
i.e.
(
0
,
−
1
)
and midpoint of
B
D
is
(
2
−
2
2
,
−
2
+
0
2
i.e.
(
0
,
−
1
)
i.e. midpoints of
A
C
and
B
D
are same,
but,
B
C
=
√
(
2
−
(
−
1
)
)
2
+
(
−
2
−
(
−
4
)
)
2
=
√
13
i.e.
A
B
≠
B
C
and hence ABCD is a parallelogram.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+1)^2+(y+4)^2-0.08)((x+2)^2+y^2-0.08)=0 [-10, 10, -5, 5]}
Case D -
C
(
1
,
−
6
)
,
D
(
0
,
−
2
)
As mid point of
A
C
is
(
1
+
1
2
,
2
−
6
2
)
i.e.
(
1
,
−
2
)
and midpoint of
B
D
is
(
2
+
0
2
,
−
2
+
(
−
2
)
2
i.e.
(
1
,
−
2
)
i.e. midpoints of
A
C
and
B
D
are same,
and,
B
C
=
√
(
2
−
1
)
2
+
(
−
2
−
(
−
6
)
)
2
=
√
17
i.e.
A
B
=
B
C
and hence ABCD is a rhombus.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-1)^2+(y+6)^2-0.08)(x^2+(y+2)^2-0.08)=0 [-14, 14, -7, 7]}
