Just do 16870 x .03 (3% as a decimal) and you'll get 506.10. Subtract that to 16870 and you'll get 16363.90.
What is the outlier in the following data set:<br>
15,11,10,8,9,1,8,7,5,4,2,3, and 37?
NeTakaya
Step-by-step explanation:
The steps to find an outlier:
1. Put the data in numerical order.
2. Find the median.
3. Find the medians for the top and bottom parts of the data. This divides the data into 4 equal parts.
The median with the smallest value is called Q1. The median for all the values - usually just called the median is also called Q2. The median with the largest value is Q3.
4. Subtract...Q3 - Q1. This value is the InterQuartileRange or IQR. Remember that the range means taking the largest minus the smallest. This is a special range having to do with the quartiles.
5. Multiply...1.5 * IQR
6. Take your answer from #5 and do 2 things with it. A). Subtract it from Q1 and B) Additional to Q3.
7. Look at all your data points. If any are SMALLER than Q1 - 1.5 *IQR, they are outliers. If any are LARGER than Q3 + 1.5 *IQR, they are also outliers.
For your data....the median, Q2 is
(43+38)/2 = 40.5.
Q1 = (30+26)/2 = 28.
Q3 = (54+52)/2 = 53
The IQR is 53 - 28 = 25
1.5 * IQR = 37.5
Q1 - 37.5 = 28 - 37.5 = -9.5. There is no data value less than -9.5.
Q3 + 37.5 = 53 + 37.5 = 90.5. there is no data value greater than 90.5.
My conclusion is that there are no outliers in this data.
I hope this helps!
Hence option a I believe is true
7 - 2/ b < 5/b
or, 7 < 5/b + 2/b
or, 7 < 7/b
divide both side by 7
1 < 1/b
or, b < 1
hope i did it correct
