Let 'a' be the number of ounces of 2%-solution in the 25-ounce mixture
and 'b' be the number of ounces of 5%-solution in the 25-ounce mixture.
Since, fluid ounces of each concentration should be combined to make 25 fl oz.
So, a+b=25 (Equation 1)
And, a container of 2% acid solution and a container of 5% acid solution should be combined to make 25 fl oz of 3.2% acid solution.
So, a of 2% + b of 5% = 3.2% of 25
Multiplying the above equation by 100, we get
(Equation 2)
Substituting the value of a=25-b in equation 2, we get
Since, a=25-b
a= 25-10
a=15.
So, 15 fluid ounces of 2% solution combined with 10 ounces of the 5% solution to create a 25-ounce mixture at 3.2% concentration of acid.
Answer:
Step-by-step explanation:
Given that a dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population of cows. She randomly assigns nold = 8 to get the old diet, and nnew = 5 to get the new diet including the additive.
From the data given we get the following
N Mean StDev SE Mean
Sample 1 8 43 5.1824 1.832
Sample 2 5 73 21.0832 9.429
df = 11
Std dev for difference = 13.3689
a) Yes the two are independent. The two sets of cows randomly chosen are definitely independent. Paired means equal number should be there and homogeneous conditions should be maintained.
b) Enclosed
c) Comparison of two means is the test recommended here. Because independent samples are used.\
d) Test statistic= -3.1233
(because of unequal variances we use that method)
95% confidence interval = ( -56.6676 , -3.3324 )
p value <0.05 our alpha
So reject null hypothesis.
The two means are statistically significantly different.
Answer: There are no fractions listed, so I can't help you. 1/3 is a repeating though, if that helps anything. Otherwise, you could use a calculator and just find the decimal of the fraction.
So you write into the table that x = 1 then y will equal 1*1.5 which equal to 1.5. So under x = 1 in y table you write y = 1.5, and so on
<span>The ratio of 3 to 4 can be written in all the following ways except 4/3. </span>
