Answer:
the answer is 3
Step-by-step explanation:
it just is
Answer:
Step-by-step explanation:
P=2(a+b)
2(26+25)
=102 ans
Sum of the numbers in the set: 42+37+32+29+20 =160
Current mean: 160/5 = 32
Median = the valvue of the middle = 32.
New mean: 32+10= 42.
Sum of numbers in the new set = 42*8 = 336
Difference: 336 - 160 = 176
I want to include 32, so that the new median stays in 32.
So the other two numbers must add 176 - 32 = 144
I will use a smaller number than 32 and the other greater (again in order to keep the same median.
I will choose 28 and 144 - 28 = 116.
So my three new numbers are 28, 32 and 116 and the new set is {116, 42, 37, 32, 32, 29,28, 20}
Checking:
Sum of the terms: 116+42+37+32+32+29+28+20 = 336
Mean = 336 / 8 = 42, which is 10 more than the original mean.
And the new median is (32+32)/2 = 32. The same of the original set.
Answer:
Step-by-step explanation:
a) about the line y = 3
⇒ 
is the intersection point
So,
b) about the line x = 5
⇒ 
So,
![V = \int\limits^3_0\pi([5-0]^2-[5-y^2/9]^2)\:dy=\pi\int\limits^3_0(25-25+10y^2/9-y^4/81)\:dy=\\\\=\pi(10y^3/27-y^5/405)|^3_0=\pi(10-3/5)=\frac{47}{5} \pi](https://tex.z-dn.net/?f=V%20%3D%20%5Cint%5Climits%5E3_0%5Cpi%28%5B5-0%5D%5E2-%5B5-y%5E2%2F9%5D%5E2%29%5C%3Ady%3D%5Cpi%5Cint%5Climits%5E3_0%2825-25%2B10y%5E2%2F9-y%5E4%2F81%29%5C%3Ady%3D%5C%5C%5C%5C%3D%5Cpi%2810y%5E3%2F27-y%5E5%2F405%29%7C%5E3_0%3D%5Cpi%2810-3%2F5%29%3D%5Cfrac%7B47%7D%7B5%7D%20%5Cpi)
