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.
<span>Let
CP ------> cost price
SP ------> Selling price
we know that
SP= CP + 0.45CP
Mark up = 0.45CP
Ratio of Mark up to Selling price-------> 0.45CP/(CP + 0.45CP)
= 0.45/(1+0.45)-------> </span><span>0.45/1.45=0.3103
</span><span>0.3103 multiplied by 100 = 31.03%
</span>
the answer is
31.03%
Let x be the number of members in Daniel's tennis team.
The amount made = $582.85
Expenses (food and drinks) = $64
Balance remaining for sharing = 582.85 - 64 = $518.85
Amount received by each player = (amount made - Expenses)/Number of members = 518.85/x
Answer:
Yes. Juan will break even because 16 and -16 are a zero pair. -16 is the additive inverse of 16. So the sum will be 0.
