Answer:
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
Step-by-step explanation:
Given that;
the frequencies of there alternatives are;
Frequency A = 60
Frequency B = 12
Frequency C = 48
Total = 60 + 12 + 48 = 120
Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;
Relative Frequency A = Frequency A / total = 60 / 120 = 0.5
Relative Frequency B = Frequency B / total = 12 / 120 = 0.1
Relative Frequency C = Frequency C / total = 48 / 120 = 0.4
therefore;
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
5000-850=
$4150 left to raise
$4150/83 students=
$50/student
Answer:
n=6
Step-by-step explanation:
12-6=6
For this case we must simplify the following expression:
![\sqrt [3] {64 * a ^ 6 * b ^ 7 * c ^ 9}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B64%20%2A%20a%20%5E%206%20%2A%20b%20%5E%207%20%2A%20c%20%5E%209%7D)
We rewrite:

So:
![\sqrt [3] {4 ^ 3 * (a ^ 2) ^ 3 * (b ^ 2) ^ 3 * b * (c ^ 3) ^ 3} =\\\sqrt [3] {4 * a ^ 2 * b ^ 2 * c ^ 3) ^ 3 * b} =](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B4%20%5E%203%20%2A%20%28a%20%5E%202%29%20%5E%203%20%2A%20%28b%20%5E%202%29%20%5E%203%20%2A%20b%20%2A%20%28c%20%5E%203%29%20%5E%203%7D%20%3D%5C%5C%5Csqrt%20%5B3%5D%20%7B4%20%2A%20a%20%5E%202%20%2A%20b%20%5E%202%20%2A%20c%20%5E%203%29%20%5E%203%20%2A%20b%7D%20%3D)
By definition of properties of powers and roots we have:
![\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20m%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bm%7D%20%7Bn%7D%7D)
So:
![4a ^ 2b ^ 2 c ^ 3 \sqrt [3] {b}](https://tex.z-dn.net/?f=4a%20%5E%202b%20%5E%202%20c%20%5E%203%20%5Csqrt%20%5B3%5D%20%7Bb%7D)
Answer:
Option B
First you should set up fractions equal to each other to see the size ratio
8/34 = x/1.2
As to make comparable fractions. Then you would cross multiply in order to get

Divide both sides by 34 and you should get an answer of 0.28m for the model beam.