Answer:
n > p + 1 + 7k
Step-by-step explanation:
21k - 3n + 9 > 3p + 12
21k - 3n > 3p + 12 - 9
3n > 3p + 3
3n > 3p + 3 + 21k
n > (3p + 3 + 21k)/3
n > p + 1 + 7k
Answer:
Step-by-step explanation:
x = 1.0
how:
we know that the Hyp of the small triangle is ![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
since we know that , then
= 1^2 + 1^2 by Pythagoras theorem
so each side is 1 :)
Answer:
a) To determine the minimum sample size we need to use the formula shown in the picture 1.
E is the margin of error, which is the distance from the limits to the middle (the mean) of the confidence interval. This means that we have to divide the range of the interval by 2 to find this distance.
E = 0.5/2 = 0.25
Now we apply the formula
n = (1.645*0.80/0.25)^2 = 27.7 = 28
The minimum sample size would be 28.
b) To answer the question we are going to make a 90% confidence interval. The formula is:
(μ - E, μ + E)
μ is the mean which is 127. The formula for E is shown in the picture.
E = 0.80*1.645/√8 = 0.47
(126.5, 127.5)
This means that the true mean is going to be contained in this interval 90% of the time. This is why it doesn't seem possible that the population mean is exactly 128.
Answer:
A)2100 ml
Step-by-step explanation:
Given total vinegar is 3 1/2 liters and the used vinegar is 3/5 of the vinegar in the beaker:
#First calculate the amount of vinegar not poured in the beaker:
![V_o=V_t-V_b\\\\=3\frac{1}{2}(1-\frac{2}{3})\\\\=1\frac{1}{6}](https://tex.z-dn.net/?f=V_o%3DV_t-V_b%5C%5C%5C%5C%3D3%5Cfrac%7B1%7D%7B2%7D%281-%5Cfrac%7B2%7D%7B3%7D%29%5C%5C%5C%5C%3D1%5Cfrac%7B1%7D%7B6%7D)
#Calculate amount of vinegar used in experiment:
![V_b=\frac{2}{3}V_t=\frac{2}{3}(3\frac{1}{2})=2\frac{1}{3}\\\\V_u=\frac{3}{5}V_b=\frac{3}{5}\times 2\frac{1}{3}=1\frac{2}{5}](https://tex.z-dn.net/?f=V_b%3D%5Cfrac%7B2%7D%7B3%7DV_t%3D%5Cfrac%7B2%7D%7B3%7D%283%5Cfrac%7B1%7D%7B2%7D%29%3D2%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5CV_u%3D%5Cfrac%7B3%7D%7B5%7DV_b%3D%5Cfrac%7B3%7D%7B5%7D%5Ctimes%202%5Cfrac%7B1%7D%7B3%7D%3D1%5Cfrac%7B2%7D%7B5%7D)
#The unused vinegar is therefore calculated by subtracting the used vinegar from the total at the start of the experiment:
![V_r=V_t-V_u\\\\=3\frac{1}{2}-1\frac{2}5}\\\\=2\frac{1}{10}\times 1000\ ml\\\\=2100\ ml](https://tex.z-dn.net/?f=V_r%3DV_t-V_u%5C%5C%5C%5C%3D3%5Cfrac%7B1%7D%7B2%7D-1%5Cfrac%7B2%7D5%7D%5C%5C%5C%5C%3D2%5Cfrac%7B1%7D%7B10%7D%5Ctimes%201000%5C%20ml%5C%5C%5C%5C%3D2100%5C%20ml)
Hence, the unused vinegar is 2100 ml