Answer:
A = 0.8 litres
B = 0.7 litres
C = 0.5 litres
D = 0.2 litres
Step-by-step explanation
Here's what we know:
1. Jug A = B + .1 litres
2. Jug C = B - 200 (or 0.2 litres)
3. Jug D = .25 x A
4. Jug A + Jug B = 1.5 litres
In problem 1, we learned that Jug A has .1 litres more than Jug B and in problem 4, the two of them added together are 1.5 litres. To solve this we can combine the problems.
B + .1 litres + B = 1.5 litres
2B + .1 = 1.5
Subtract .01 from each side and you have 2B = 1.4
Divide each side by 2 and you have B = 0.7 litres
Plug this info into problem 1 and you can solve for A. (0.7 + 0.1 = 0.8)
Plug this info into problem 2 and you can solve for C. (0.7 - 0.2 = 0.5)
Since you have A, you can use that info to solve problem 3 (0.25 x 0.8 = 0.2)
Answer:
0.5
Step-by-step explanation:
I found this out because I am smart
Solution
Problem 6
For this case we can do this:
12, 16,__, 14, 8, 7
We can solve for x like this:
Problem 7
F
Answer:
We have strong evidence that on average, students study less than 150 minutes per night during the school week
Step-by-step explanation:
Normal distribution:
mean μ₀ = 150
Sample:
Sample size n = 272
Sample mean x = 141
Sample standard deviation s = 66
The standard error of the sample mean SE = σ /√n
SE = 66/√272
SE = 66 / 16,49
SE = 4
Test Hypothesis:
Null hypothesis H₀ x = μ₀
Alternative hypothesis Hₐ x < μ₀
z(s) test statistics is:
z(s) = ( x - μ₀ ) / s/√n
z(s) = - 9 /4
z(s) = - 2,25
p-value for that z(s) p-value = 0,0122
Then for α = 0,05 p-value < 0,05
We are in the rejection region we need to reject H₀
