Answer:
We fail to reject H0; Hence, we conclude that there is no significant evidence that the mean amount of water per gallon is different from 1.0 gallon
Pvalue = - 2
(0.98626 ; 1.00174)
Since, 1.0 exist within the confidence interval, then we can conclude that mean amount of water per gallon is 1.0 gallon.
Step-by-step explanation:
H0 : μ= 1
H1 : μ < 1
The test statistic :
(xbar - μ) / (s / sqrt(n))
(0.994 - 1) / (0.03/sqrt(100))
-0.006 / 0.003
= - 2
The Pvalue :
Pvalue form Test statistic :
P(Z < - 2) = 0.02275
At α = 0.01
Pvalue > 0.01 ; Hence, we fail to reject H0.
The confidence interval :
Xbar ± Margin of error
Margin of Error = Zcritical * s/sqrt(n)
Zcritical at 99% confidence level = 2.58
Margin of Error = 2.58 * 0.03/sqrt(100) = 0.00774
Confidence interval :
0.994 ± 0.00774
Lower boundary = (0.994 - 0.00774) = 0.98626
Upper boundary = (0.994 + 0.00774) = 1.00174
(0.98626 ; 1.00174)
Answer:
Solution below.
Step-by-step explanation:
This question may seem complicated, but it is really on the concept of substitution.
So given:
f(x) = 3x
g(x) = x - 4
h(x) =
We are asked to find (h•(f-g))(2).
Note: This is not a composite function.
First, lets find (h•(f-g))(x) first.
(f-g)(x) = f(x) - g(x) = 3x - (x - 4) = 2x + 4
(h•(f-g))(x) =
Now we can find (h•(f-g))(2) by substituting x = 2 into the function.
(h•(f-g))(2) =
Answer:
Step-by-step explanation: The remainder is 200
Answer:
No she is wrong
Step-by-step explanation:
The correct answer is 320
Answer:
14.59%.
Step-by-step explanation:
Part A
You can not accurately calculate this because the SRS is not over 30. The CLT states that if the sample size is big enough, it will have a normal distribution. 15 is not big enough.
Part B
You can calculate this because the sample size is over 30; in this case, it is 40. So first, we have to find the standard deviation of the sampling distribution of the means. Which is done by taking the standard deviation and dividing it by the square root of the sample size, which comes out to be 4.74341649. Next, we throw it into the calculator in Normal CDF (260,9999,255,4.74341649). The final answer comes out to be 14.59%.