Answer:
B. Oliver can watch TV for 1.5 hours more this week.
Step-by-step explanation:
Oliver is not allowed to watch more than 4 hours of television a week. He watched his favorite show on Monday which was 1 hour long and his favorite Tuesday show which was 1.5 hours long.
So he has already finished 2.5 hours of watching . He can watch only for 4 hours for the total week.
Let x be the number of more hours for which he can watch television.
Then ,
x + 1 + 1.5 = 4
x + 2.5 = 4
x = 4 - 2.5
x = 1.5 hours.
So Oliver can watch television for at most 1.5 hours for the rest of the week.
Then the correct option is B.
Answer:
12
Step-by-step explanation:
Its 12 because when you do -16 + - 40.... then -12 you get =12 There you go
FACTORISATION IS BASICALLY FINDING LIKE-TERMS.
1. FIND THE HCF OF 8. HERE, THE HCF IS 8. ALSO, CHECK ALL THE LIKE-TERMS. LIKE-TERMS ARE OUTSIDE THE BRACKET AND UNLIKE TERMS ARE INSIDE.
2. ADD THE TERMS IN THE BRACKET AND THE TERMS OUTSIDE THE BRACKET FROM STEP 1.
8x + 8y + rx + ry
1. 8 (x + y) + r (x + y)
2. (x +y) (8+r)
(side 1)² + (side 2)² = (diagonal)²
(9.5² + (14)² = (diagonal)²
90.25 + 196 = (diagonal)²
286.25 = (diagonal)²
√286.25 = √(diagonal)²
17 = diagonal
Answer: 17 inches
Answer:
The value of the constant C is 0.01 .
Step-by-step explanation:
Given:
Suppose X, Y, and Z are random variables with the joint density function,
The value of constant C can be obtained as:
![C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1](https://tex.z-dn.net/?f=C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5Cint%5Climits%5E%5Cinfty_0%7Be%5E%7B-0.2y%7D%28%5B%5Cfrac%7B-e%5E%7B-0.1z%7D%20%7D%7B0.1%7D%20%5D%5Climits%5E%5Cinfty__0%20%7D%29%20%5C%2C%20dy%20%20%7D%29%20%5C%2C%20dx%20%3D%201)
![C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1](https://tex.z-dn.net/?f=C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.2y%7D%28%5B%5Cfrac%7B-e%5E%7B-0.1%28%5Cinfty%29%7D%20%7D%7B0.1%7D%2B%5Cfrac%7Be%5E%7B-0.1%280%29%7D%20%7D%7B0.1%7D%20%5D%29%20%20%7D%20%5C%2C%20dy%20%20%7D%29%20%5C%2C%20dx%20%3D%201)
![C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1](https://tex.z-dn.net/?f=C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.2y%7D%5B0%2B%5Cfrac%7B1%7D%7B0.1%7D%5D%20%20%7D%20%5C%2C%20dy%20%20%7D%29%20%5C%2C%20dx%20%3D1)
![10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1](https://tex.z-dn.net/?f=10C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5B%5Cfrac%7B-e%5E%7B-0.2y%7D%20%7D%7B0.2%7D%5D%5E%5Cinfty__0%20%20%7D%29%20%5C%2C%20dx%20%3D%201)
![10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}] } \, dx = 1](https://tex.z-dn.net/?f=10C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5B%5Cfrac%7B-e%5E%7B-0.2%28%5Cinfty%29%7D%20%7D%7B0.2%7D%2B%5Cfrac%7Be%5E%7B-0.2%280%29%7D%20%7D%7B0.2%7D%5D%20%20%20%7D%20%5C%2C%20dx%20%3D%201)
![10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1](https://tex.z-dn.net/?f=10C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%5B0%2B%5Cfrac%7B1%7D%7B0.2%7D%5D%20%20%7D%20%5C%2C%20dx%20%3D%201)
![50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1](https://tex.z-dn.net/?f=50C%28%5B%5Cfrac%7B-e%5E%7B-0.5x%7D%20%7D%7B0.5%7D%5D%5E%5Cinfty__0%7D%29%20%3D%201)
![50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1](https://tex.z-dn.net/?f=50C%5B%5Cfrac%7B-e%5E%7B-0.5%28%5Cinfty%29%7D%20%7D%7B0.5%7D%20%2B%20%5Cfrac%7B-0.5%280%29%7D%7B0.5%7D%5D%20%3D1)
![50C[0+\frac{1}{0.5} ] =1](https://tex.z-dn.net/?f=50C%5B0%2B%5Cfrac%7B1%7D%7B0.5%7D%20%5D%20%3D1)
⇒ 
C = 0.01
