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: wait
Step-by-step explanation:
7/9=35/45
8/15=24/45
35/45÷24/45=
<span>7/9</span>÷<span>8/15</span>=<span>7/9</span>×<span>15/8</span>=<span>7</span>×<span> 15/9 </span>×<span> 8</span>=<span>10572</span>=<span>35 </span>×<span> 324 </span>×<span> 3</span>=<span>3524</span>=1 <span>11/24</span><span>
I hope this helped!!!:)
</span>
Answer:
C. (-3, 6.5)
Step-by-step explanation:
Hope it helps.