Answer:
48 ounces of graphene covers 448 acres.
Step-by-step explanation:
One ounce of graphene cover football fields = 7
Area of 1 football field =![1 \frac{1}{3} = \frac{4}{3} acres](https://tex.z-dn.net/?f=1%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20acres)
Area of 7 football fields = ![\frac{4}{3} \times 7 = \frac{28}{3} acres](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%207%20%3D%20%5Cfrac%7B28%7D%7B3%7D%20acres)
So, 1 ounce of graphene cover acres of field = ![\frac{28}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B28%7D%7B3%7D)
48 ounces of graphene cover acres of field = ![\frac{28}{3} \times 48 = 448 acres](https://tex.z-dn.net/?f=%5Cfrac%7B28%7D%7B3%7D%20%5Ctimes%2048%20%3D%20448%20acres)
Hence 48 ounces of graphene covers 448 acres.
Standard deviation=4.
let the mean be 12 and the score be 0 (therefore 12 point difference.)
0-12/s=-3 (s being standard deviation)
-12/4 = -3
Answer:
-5
Step-by-step explanation:
counting is like backwards in negatives so a +5 is slow and a +25 is high but in negatives its the opposite because you are missing that many
I dont really know who to explain it but I hope this helps
R
2
=
9.61
O
h
m
s
Explanation:
l
1
=
18
m
r
1
=
1.2
2
=
0.6
m
m
=
6
⋅
10
−
4
m
R
1
=
10
O
h
m
s
R
1
=
τ
⋅
l
π
⋅
r
2
1
R
1
=
τ
⋅
18
π
⋅
36
⋅
10
−
8
10
=
τ
⋅
18
36
⋅
π
⋅
10
−
8
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
l
2
=
27
m
r
2
=
1
⋅
5
2
=
0.75
⋅
10
−
3
=
7.5
⋅
10
−
4
m
R
2
=
τ
⋅
l
2
π
⋅
r
2
2
R
2
=
τ
⋅
27
π
⋅
56.2
⋅
10
−
8
R
2
=
τ
⋅
27
56.2
π
⋅
10
−
8
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
.
.
10
R
2
=
τ
⋅
18
36
⋅
π
⋅
10
−
8
⋅
56.2
⋅
π
⋅
10
−
8
τ
⋅
27
10
R
2
=
18
⋅
56.2
36
⋅
27
10
R
2
=
56.2
54
R
2
=
540
56.2
R
2
=
9.61
O
h
m
s