To determine the average water use, 109 people must be sampled.
1.9 gallons is the standard deviation.
E = 0.15 gallons maximum error
1.9 gallons on average
90% is the critical value.
1.645 is the 90% confidence interval.
sample size requirement,
n = (((z ÷ 2) × σ) ÷ E)²
n = (((1.645 ÷ 2) × 1.9) ÷ 0.15)²
n = ((0.8225 × 1.9) ÷ 0.15)²
n = (1.5627 ÷ 0.15)²
n = (10.418)²
n = 108.53 ≈ 109
As a result, the minimal sample size necessary to determine the mean water use = 109.
It is determined by dividing the average standard error by the squared of the sample size, and it decreases with increasing sample size. In other words, when the sample size is sufficiently big, the population mean approaches the population mean.
To learn more about the sample interval at
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Answer:
m = 13.
Step-by-step explanation:
Recall that adding logarithmic equations, (ex: log a + log b) is the same as log (a·b). Therefore:
log 12 + log 5 = log (4m + 8)
log (12·5) = log (4m + 8)
log (60) = log (4m + 8) **Ignore 'Log' to solve for 'm'
60 = 4m + 8 **Subtract both sides by 8
52 = 4m **Divide both sides by 4
m = 13.
It's 96 pages that it would print in one hour
Answer:
6d^5 - 3c^3d^2 + 5c^2d^3 + c^3d^4 - 12cd^4 + 8
Step-by-step explanation:
We need to subtract the given polynomial from the sum:-
8d^5 - 3c^3d^2 + 5c^2d^3 - 4cd^4 + 9 - (2d^5 - c^3d^4 + 8cd^4 +1 )
We need to distribute the negative over the parentheses:-
= 8d^5 - 3c^3d^2 + 5c^2d^3 - 4cd^4 + 9 - 2d^5 + c^3d^4 - 8cd^4 -1
Bringing like terms together:
= 8d^5 - 2d^5 - 3c^3d^2 + 5c^2d^3 + c^3d^4 - 4cd^4 - 8cd^4 + 9
- 1
Simplifying like terms
= 6d^5 - 3c^3d^2 + 5c^2d^3 + c^3d^4 - 12cd^4 + 8
=2(5x2+20x-1)
The 2 by the x has to be on top of the x
