3xy
<span>y(3y)/3xy + y(xy)/3xy + (y+1)(3x)/3xy </span>
<span>NOW since all of the fractions have a denominator of 3xy, drop the denominators and solve using the numerators. </span>
<span>y(3y) + y(xy) + (y+1)(3x) </span>
<span>3y^2 + xy^2 + 3xy +3x </span>
<span>cannot simplify further.</span>
Answer: thanks for the points
Step-by-step explanation:
(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes
(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)
Variance = Standard deviation ^2
Therefore,
Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes
(c) Probability that it will take more than 30 minutes to get to school
P(x>30) = 1-P(x=30)
Z(x=30) = (mean-30)/SD = (28-30)/4.47 ≈ -0.45
Now, P(x=30) = P(Z=-0.45) = 0.3264
Therefore,
P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%
With actual average time to walk to the bus stop being 10 minutes;
(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes
(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.
(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0
From Z table, P(x=30) = 0.5
And therefore, P(x>30) = 1- P(X=30) = 1- P(Z=0.0) = 1-0.5 = 0.5 = 50%
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
Step-by-step explanation:
first you have to take all similar variable on the same side of the equation and the constants on the other side then solve the question .
example : -12x + 9 = -15x
-12x +15x = -9
3 x = -3
X = -3/3= -1
x = -1
like this you can solve all the other equation ..
<em>Hope </em><em>this </em><em>will </em><em>be</em><em> helpful</em><em> to</em><em> you</em><em> </em><em>.</em><em>.</em>
<em>plz </em><em>mark</em><em> my</em><em> answer</em><em> as</em><em> brainlist</em><em> </em><em>if </em><em>you</em><em> </em><em>find </em><em>it </em><em>useful</em><em>.</em><em>.</em>
