The answer is c!!!!!!!!!!!!!!!!!!!!!!!!
Let x represent the number of liters of 50% acid Theresa puts into the mix. The the number of liters of 30% acid will be (420-x). The total amount of acid in the final solution will be ...
0.50x + 0.30(420-x) = 0.45(420)
0.20x + 126 = 189 . . . . . . . . . . . . . . . simplify
0.20x = 63 . . . . . . . . . . . . . . . . . . . . . subtract 126
x = 63/0.20 = 315 . . . . . . . . . . . . . . . liters of 50% solution
(420-x) = 420-315 = 105 . . . . . . . . . liters of 30% solution
Theresa should mix ...
105 liters of 30% solution
315 liters of 50% solution
Answer:
Yes. The data provide enough evidence to support the claim that the mean weight of one-year-old boys is greater than 25 pounds.
P-value=P(t>2.84)=0.0024
Step-by-step explanation:
Hypothesis test on the population mean.
The claim is that the mean weight of one-year-old boys is greater than 25 pounds.
Then, the null and alternative hypothesis are:

The significance level is α=0.05.
The sample size is n=354. The sample mean is 25.8 pounds and the sample standard deviation is 5.3 pounds. As the population standard deviation is estimated from the sample standard deviation, we will use a t-statistic.
The degrees of freedom are:

The t-statistic is:

For a right tailed test and 353 degrees of freedom, the P-value is:

As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected.
There is enough evidence to support the claim that the mean weight of one-year-old boys is greater than 25 pounds.
Q1: 7×6 = 42
Q2: 42×1/8 = 42/8 = 5.25 (5 1/4) in^3
Q3: (0.5)^3 = 0.5×0.5×0.5 = 0.125 m^3
Q4: 4(5/2)^2 = 4×5/2×5/2 = 4×25/4 = 25 in^3
Q5: 3(13/2)(3/2) = 117/4 = 29.25 (29 1/4) in^3
Q6: (0.9)^3 = 0.729 cm^3
The answer to your problem is 2