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
Answer: B
Step-by-step explanation:
I got it right on Edge
Answer:
4.5 % alloy
Step-by-step explanation:
Let x = amt of 30% alloy
The resulting total is to be 25 oz, therefore:
(25-x) = amt of 5% alloy:
A typical mixture equation:
.30x + .05(25-x) = .20(25)
.30x + 1.25 - .05x = 5 .30x - .05x = 5 - 1.25
.25x = 3.75
x = 3.75%2F.25
x = 15 oz of 30% alloy required
then
25-15 = 10 oz of 5% alloy:
Check solution
.30(15) + .05(10) = .20(25)
4.5 + .5 = 5
Answer:
4
Step-by-step explanation:
2+2=4
