Answer:
option (b) 0.0228
Step-by-step explanation:
Data provided in the question:
Sample size, n = 100
Sample mean, μ = $20
Standard deviation, s = $5
Confidence interval = between $19 and $21
Now,
Confidence interval = μ ± ![z\frac{s}{\sqrt n}](https://tex.z-dn.net/?f=z%5Cfrac%7Bs%7D%7B%5Csqrt%20n%7D)
thus,
Upper limit of the Confidence interval = μ + ![z\frac{s}{\sqrt n}](https://tex.z-dn.net/?f=z%5Cfrac%7Bs%7D%7B%5Csqrt%20n%7D)
or
$21 = $20 + ![z\frac{5}{\sqrt{100}}](https://tex.z-dn.net/?f=z%5Cfrac%7B5%7D%7B%5Csqrt%7B100%7D%7D)
or
z = 2
Now,
P(z = 2) = 0.02275 [From standard z vs p value table]
or
P(z = 2) ≈ 0.0228
Hence,
the correct answer is option (b) 0.0228
Answer:
2 698 132 122 m bytes / s
Step-by-step explanation:
Answer:
Don't put it on the news because it's a high risk personnel?
Answer:
We can conclude that the setting is stationed somewhere in the north where it's very cold, so cold that they can use sled dogs. We can also assume it's winter time because it normally snows during the winter.
So the setting is in the north where it is cold, during the winter time.
Answer:
The rate of the volume increase will be ![\frac{dV}{dt}=50.27 cm^{3}/s](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D50.27%20cm%5E%7B3%7D%2Fs)
Step-by-step explanation:
Let's take the derivative with respect to time on each side of the volume equation.
![\frac{dV}{dt}=4\pi R^{2}\frac{dR}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%20R%5E%7B2%7D%5Cfrac%7BdR%7D%7Bdt%7D)
Now, we just need to put all the values on the rate equation.
We know that:
dR/dt is 0.04 cm/s
And we need to know what is dV/dt when R = 10 cm.
Therefore using the equation of the volume rate:
![\frac{dV}{dt}=4\pi 10^{2}0.04](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%2010%5E%7B2%7D0.04)
![\frac{dV}{dt}=50.27 cm^{3}/s](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D50.27%20cm%5E%7B3%7D%2Fs)
I hope it helps you!