1 minute = 60 seconds.
Multiply total minutes by 60 to get total seconds:
1.8 minutes x 60 seconds per minute = 108 seconds.
1 minute = 1/60th of an hour:
1.8 minutes x 1/60 = 0.03 hours
Answer: x=2
Step-by-step explanation:
-4x+7=-1
-4x=-8
x=2
Answer:
The approximate standard deviation of the sampling distribution of the mean for all samples of size n is ![s = \frac{\sigma}{\sqrt{n}} = \frac{21}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cfrac%7B21%7D%7B%5Csqrt%7Bn%7D%7D)
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The mean life expectancy of a certain type of light bulb is 945 hours with a standard deviation of 21 hours
This means that
.
What is the approximate standard deviation of the sampling distribution of the mean for all samples of size n?
![s = \frac{\sigma}{\sqrt{n}} = \frac{21}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cfrac%7B21%7D%7B%5Csqrt%7Bn%7D%7D)
The approximate standard deviation of the sampling distribution of the mean for all samples of size n is ![s = \frac{\sigma}{\sqrt{n}} = \frac{21}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cfrac%7B21%7D%7B%5Csqrt%7Bn%7D%7D)
Answer:
2.3 by 10^7
Step-by-step explanation:
We want this rewritten in the form 2.3*10^n.
To get the decimal point in 23,000,000 between the 2 and 3, we must move it 7 places to the left, and then compensate for this by multiplying 2.3 by 10^7.
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