Answer:
0.3
Step-by-step explanation:
Answer:
x = 5
Step-by-step explanation:
A = hb/2
10 = 4(x)/2
10 = 2x
x = 5
im not sure if this is right but get back to me if it isn't
Bror cancelled terms which should NEVER be done.
x² + 3x + 2 / (x + 2) =
(x+1) * (x+2) / (x+2)
Cancelling the (x+2) factor we get
x + 1
Answer:
The correct answer is D. 0.77
Step-by-step explanation:


![=Pr[\frac{-1}{4}\leq \bar{X}_{48}\leq \frac{1}{4}]\\\\=Pr[\frac{-0.25}{0.2083}\leq Z\leq \frac{0.25}{0.2083}]\\\\=Pr[-1.2\leq Z\leq 1.2]\\=Pr(Z\leq 1.2)-Pr(Z\leq -1.2)\\=Pr(Z\leq 1.2)-(1-Pr(Z\leq 1.2))\\=2\times 0.8849 -1\text{ ( Z value for 1.2 is 0.8849 )}\\\approx 0.77](https://tex.z-dn.net/?f=%3DPr%5B%5Cfrac%7B-1%7D%7B4%7D%5Cleq%20%5Cbar%7BX%7D_%7B48%7D%5Cleq%20%5Cfrac%7B1%7D%7B4%7D%5D%5C%5C%5C%5C%3DPr%5B%5Cfrac%7B-0.25%7D%7B0.2083%7D%5Cleq%20Z%5Cleq%20%5Cfrac%7B0.25%7D%7B0.2083%7D%5D%5C%5C%5C%5C%3DPr%5B-1.2%5Cleq%20Z%5Cleq%201.2%5D%5C%5C%3DPr%28Z%5Cleq%201.2%29-Pr%28Z%5Cleq%20-1.2%29%5C%5C%3DPr%28Z%5Cleq%201.2%29-%281-Pr%28Z%5Cleq%201.2%29%29%5C%5C%3D2%5Ctimes%200.8849%20-1%5Ctext%7B%20%28%20Z%20value%20for%201.2%20is%200.8849%20%29%7D%5C%5C%5Capprox%200.77)
Hence, the correct answer is 0.77
Answer:
It will take 11/30(cistern/hour) for the tank to be full