C) 30 * IQR is the box. So the highest point on the box is 95 and the lowest is 65, which makes the difference 30. I hope this helped!!
Answer: (A) 
(B) Length varies between 1 and 150
(C) Largest area is 22500ft²
Step-by-step explanation: Suppose length is l and width is w.
The rectangular garden has perimeter of 600ft, which is mathematically represented as
Area of a rectangle is calculated as
Now, we have a system of equations:
Isolate w, so we have l:
w = 300 - l
Substitute in the area equation:
A = l(300 - l)
A = 300l - l²
(A) <u>Function of area in terms of length is given by </u><u>A = 300l - l²</u>
(B) The practical domain for this function is values between 1 and 150.
(C) For the largest area, we need to determine the largest garden possible. For that, we take first derivative of the function:
A' = 300 - 2l
Find the values of l when A'=0:
300 - 2l = 0
2l = 300
l = 150
Replace l in the equation:
w = 300 - 150
w = 150
Now, calculate the largest area:
A = 150*150
A = 22500
<u>The largest area the fence can enclose is </u><u>22500ft².</u>
Answer:
Confidence interval variance [21.297 ; 64.493]
Confidence interval standard deviation;
4.615, 8.031
Step-by-step explanation:
Given :
Variance, s² = 34.34
Standard deviation, s = 5.86
Sample size, n = 27
Degree of freedom, df = 27 - 1 = 26
Using the relation for the confidence interval :
[s²(n - 1) / X²α/2, n-1] ; [s²(n - 1) / X²1-α/2, n-1]
From the chi distribition table :
X²α/2, n-1 = 41.923 ; X²1-α/2, n-1 = 13.844
Hence,
[34.34*26 / 41.923] ; [34.34*26 / 13.844]
[21.297 ; 64.493]
The 95% confidence interval for the population variance is :
21.297 < σ² < 64.493
Standard deviation is the square root of variance, hence,
The 95% confidence interval for the population standard deviation is :
4.615 < σ < 8.031
Answer:
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z=1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).
