This should be simple, add the Amounts of snow for the first week
4.32 + 6.86 = 11.18
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The subtract the amounts of snow from each week
11.18 - 7.89 = 3.29
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C.) 3.29 is your answer
We can solve this problem by referring to the standard
probability distribution tables for z.
We are required to find for the number of samples given the
proportion (P = 5% = 0.05) and confidence level of 95%. This would give a value
of z equivalent to:
z = 1.96
Since the problem states that it should be within the true
proportion then p = 0.5
Now we can find for the sample size using the formula:
n = (z^2) p q /E^2
where,
<span> p = 0.5</span>
q = 1 – p = 0.5
E = estimate of 5% = 0.05
Substituting:
n = (1.96^2) 0.5 * 0.5 / 0.05^2
n = 384.16
<span>Around 385students are required.</span>
Answer:
Section A = 25,000 seats
Section B = 14,600 seats
Section C = 10,400 seats
Step-by-step explanation:
Total Seats = 50,000
Seats in Section A cost = $30
Seats in Section B cost = $24
Seats in Section C cost = $18
Total sales from the event = $1,287,600
No. of Seats in section A = No. seats in Section B + No. seats in Section C
A = B + C
or, 2A = 50,000
A = 25,000 seats @ $30/seat = $750,000
B + C = 25,000
24B + 18C = 537,600
24B + 18(25,000 - B) = 537,600
24B + 450,000 - 18B = 537,600
6B = 87600
B = 14,600
C = 10,400
Hence;
A = 25,000 seats
B = 14,600 seats
C = 10,400 seats
Answer:
8.12 is the rounded answer
Step-by-step explanation:
