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
Answer:16.05 $
Step-by-step explanation: Put all the numbers in a vertical format and add them all up
9.95
+1.15
+4.95
--------
16.05
Answer:
15 and 18
Step-by-step explanation:
33 split into groups of 5 and 6
Answer:
f (x) = x2 shows us that function " f " takes " x " and squares it. Example: with f (x) = x2: becomes an output of 16. In fact we can write f (4) = 16. The "x" is Just a Place-Holder! Don't get too concerned about "x", it is just there to show us where the input goes and what happens to it.
In order to solve this problem, you must draw a right triangle first to help you visualize the path of the jogger.
You are given 0.5 miles north (which will be the height of the right triangle), and 1.3 miles (which will be the hypotenuse of the triangle).
Using Pythagorean Theorem: a²+b²=c², we have:
0.5² + b² = 1.3²
b = 1.3²-0.5²
b = 1.2 miles
