Answer:
Step-by-step explanation:
Let chocolate bar = c and blow pops = b
<u>As per given we have equations below:</u>
- 5c + 8b = 23.25
- 2c + 13b = 21.55
<u>13 times the first equation minus 8 times the second to eliminate b and find c:</u>
- 13(5c + 8b) - 8(2c + 13b) = 13(23.25) - 8(21.55)
- 65c - 16c = 129.85
- 49c = 129.85
- c = 129.85/49
- c = 2.65
Each chocolate bar costs $2.65
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.
If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?
Given Information:
standard deviation = σ = 30 hours
confidence level = 99%
Margin of error = 6 hours
Required Information:
sample size = n = ?
Answer:
sample size = n ≈ 165
Step-by-step explanation:
We know that margin of error is given by
Margin of error = z*(σ/√n)
Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size
√n = z*σ/Margin of error
squaring both sides
n = (z*σ/Margin of error)²
For 99% confidence level the z-score is 2.576
n = (2.576*30/6)²
n = 164.73
since number of bulbs cannot be in fraction so rounding off yields
n ≈ 165
Therefore, a sample size of 165 bulbs is needed to ensure a margin of error not greater than 6 hours.
More than 10 a gumball.
150 divided by 15 is 10, which would be the price you bought for. If you want to make more money, you must at least have a surge pricing of 10 unit prices per gumball.
