Answer:
a) 20<h≤30.
b) 26.17 hrs
Step-by-step explanation:
The missing table is shown in attachment.
Part a)
We need to find the class interval that contains the median.
The total frequency is
The median class corresponds to half
That is the 15th value.
We start adding the frequency from the top obtain the least cumulative frequency greater or equal to 15.
2+8+9=19
This corresponds to the class interval 20<h≤30.
Adding from the bottom also gives the same result.
Therefore the median class is 20<h≤30.
b) Since this is a grouped data we use the midpoint to represent the class.
The median is given by :
You can solve this easily by using Pascal's Triangle (look that up if need be).
Here are the first four rows of P. Triangle:
1
1 1
1 2 1
1 3 3 1
example: expand (a+b)^3.
Look at the 4th row. Borrow and use those coefficients:
1a^3 + 3 a^2b + 3ab^2 + b^3
Now expand (4x+3y)^3:
1(4x)^3 + 3(4x)^2(3y) + 3(4x)*(3y)^2 + (3y)^3
Look at the 2nd term (above):
3(4x)^2(3y) can be re-written as 144x^2y.
The coeff of the 2nd term is 144. Note that (4)^2 = 16
290 mph , its going half the speed as before. Hope this helps.
Answer:
that does not make sense ?????????????????????????????????????????
Answer:
The cube root of given expression is 
.
Step-by-step explanation:
The given expression is
![\sqrt[3]{216x^9y^{18}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216x%5E9y%5E%7B18%7D%7D)
It can be written as
Use the exponent property: 
![\sqrt[3]{216x^9y^{18}}=\sqrt[3]{6^3(x^3)^3(y^6)^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216x%5E9y%5E%7B18%7D%7D%3D%5Csqrt%5B3%5D%7B6%5E3%28x%5E3%29%5E3%28y%5E6%29%5E%7B3%7D%7D)
![\sqrt[3]{216x^9y^{18}}=\sqrt[3]{(6x^3y^6)^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216x%5E9y%5E%7B18%7D%7D%3D%5Csqrt%5B3%5D%7B%286x%5E3y%5E6%29%5E%7B3%7D%7D)
![\sqrt[3]{216x^9y^{18}}=6x^3y^6](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216x%5E9y%5E%7B18%7D%7D%3D6x%5E3y%5E6)
Therefore the cube root of given expression is .
