Median is the middle number
60, 60, 65, 70, 70, 85, 90
70 is your median
mode is the number(s) that show up the most
60 and 70 is your mode, since they show up twice (one more than the others)
Range is largest number minus the smallest
90 - 60 = 30, 30 is your range
Mean is all the numbers added together divided by the number of numbers there are
60 + 90 + 65 + 70 + 70 + 85 + 60 = 500
500/7 = 71.42
Mean = 71.42
hope this helps
1-0/-1-0 = -1
Y- 0 = -(x-0)
Y=-x
Y - 1 = -(x+1)
Y-1= -x-1
Y = -x
Answer:
What is the Question so I can answer?
Step-by-step explanation:
To solve this problem, you'd want to start by finding the mean of the given numbers. To find the mean, add all the numbers together and divide by how many there are.
Next, you'll see that the question says one of the rents changes from $1130 to $930. So find the mean of all the numbers again, except include $930 in your calculation instead of $1130.
I got $990 as the mean for the given numbers, and $970 as the mean after replacing the $1130 with $930. Subtracting the two means gives you $20. So the mean decreased by $20.
Now for the median, all you need to do is find the median of the given numbers and compare them with the median of the new data. Because there are ten terms, you have to add the middle two numbers and divide by two. $990 + $1020 = 2010. 2010÷2 = $1005 as the first median.
The new rent is 930, so you have to reorder the data so it goes from least to greatest again. 745, 915, 925, 930, 965, 990, 1020, 1040, 1050, 1120. After finding the median again you get 977.5. Subtracting the two medians gives you $27.5 as how much the median decreased. Hope this helps!
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.
