Answer:
i think the answer is $3.45
Step-by-step explanation:
just had to divide the two numbers
Answer:
P = 0.0438 ; We reject the Null and conclude that there is significant evidence that the discharge limit per gallon of waste has been exceeded.
Due to the small sample size.
Step-by-step explanation:
H0: μ = 500
H1 : μ > 500
Test statistic :
(xbar - μ) / S.E
Tstatistic = (1000 - 500) / 200
= 500 / 200
= 2.5
Pvalue from Test score ; df = 4 - 1 = 3 ; using calculator :
Pvalue = 0.0438
At α = 0.05
Pvalue < α ; We reject the Null and conclude that there is significant evidence that the discharge limit per gallon of waste has been exceeded.
The test above has a very small sample size, and for a distribution to be approximately Normal, the sample size must be sufficiently large enough according to the Central limit theorem.
For a two sided analysis ; the Pvalue is twice that for the one sided, hence, Pvalue = (0.0438 * 2) = 0.0876 yielding a less strong evidence against the Null.
Answer:
The answer is 154.8 ml
Step-by-step explanation:
In this question, the amount of pure acid is 18% of the total solution, that is, 18% of 860 milliliters. So
0.18*860 = 154.8 ml
The answer is 154.8 ml
Would it not be 4:5 as a ratio?
Elimination
first i would divide the second equation by -2.
x - y = -2
-x- 2y = -19
the x is eliminated.
-3y = -21
divide each side by -3.
y = 7
plug 7 in for y to solve for x
x-7 = -2
x = 5
(5,7)
