Answer: Yes. The new average is 3
<u>Step-by-step explanation:</u>
25 students at 3 siblings each = 75 siblings
26 students has 75 siblings + 8 siblings = 83 siblings
New average is 83 siblings ÷ 26 students = 3.19 siblings per students
Answer:
Step-by-step explanation:
We want to find the Riemann sum for 
with n = 6, using left endpoints.
The Left Riemann Sum uses the left endpoints of a sub-interval:
where 
.
Step 1: Find 
We have that 
Therefore, 
Step 2: Divide the interval ![\left[0,\frac{3 \pi}{4}\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%5Cright%5D)
into n = 6 sub-intervals of length 
![a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b](https://tex.z-dn.net/?f=a%3D%5Cleft%5B0%2C%20%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B4%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%5Cright%5D%3Db)
Step 3: Evaluate the function at the left endpoints
Step 4: Apply the Left Riemann Sum formula
Answer:
R is a function.
Step-by-step explanation:
Answer:
Given that The data to represent average test scores for a class of 16 students includes an outlier value of 78.
We can find sum of all 16 test scores = 84(16) = 1344
Outlier found = 78
If outlier is removed new sum = 1344-78 = 1266
Number of entries without outlier = 15
New average = 1266/12 =84.4
We find that average of new data increases.
Also whenever we remove outlier std deviation also would be reduced.
Step-by-step explanation:
