Answer:
(x+9,y-12)
Step-by-step explanation:
sorry if this didnt help
Answer:40,600
Step-by-step explanation:
10x10x10x10=10,000
10x10=100
4x10,000+6x100
40,000+600=
40,600
Answer:
-6
Step-by-step explanation:
Answer:
a) 
b) 
c) 
With a frequency of 4
d) 
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:

Replacing we got:

Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Part c
The mode is the most repeated value in the sample and for this case is:

With a frequency of 4
Part d
The midrange for this case is defined as:

Part e
For this case we can calculate the deviation given by:

And replacing we got:

And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Answer:
4.75
Step-by-step explanation:
19 goes into 4 4 times, so we are left with 4 3/4. 3/4 is 0.75, so we have 4.75 as our answer.