Lol idk.......................................................................................................
Answer:
m∠P′Q′R′ = m∠PQR
Step-by-step explanation:
Let's start first by writing down the given:
σ = 100
sample mean = 450
sample size = 25
These information, plus the fact that we know that the population is approximately normally distributed, would tell us that we can use the
normal distribution curve in analyzing the problem.
A confidence interval of the mean is just a range statistically estimated to contain the population mean. For a 90% confidence interval, we would look at the Z-table and see where 90% of the data falls. We'll notice that it will fall within
1.645 standard deviations of the mean.
Next, we look for the standard error of the mean. This will have a formula

The standard error would just therefore be equal to

Lastly, we just get the product of the standard error and 1.645 and add it to 450 for the maximum value and subtract it to 450 for the minimum value.

ANSWER: 417.1<μ<482.9
Answer:
The center/ mean will almost be equal, and the variability of simulation B will be higher than the variability of simulation A.
Step-by-step explanation:
Solution
Normally, a distribution sample is mostly affected by sample size.
As a rule, sampling error decreases by half by increasing the sample size four times.
In this case, B sample is 2 times higher the A sample size.
Now, the Mean sampling error is affected and is not higher for A.
But it's sample is huge for this, Thus, they are almost equal
Variability of simulation decreases with increase in number of trials. A has less variability.
With increase number of trials, variability of simulation decreases, so A has less variability.