Answer:
(a) 3.75
(b) 2.00083
(c) 0.4898
Step-by-step explanation:
It is provided that X has a continuous uniform distribution over the interval [1.3, 6.2].
(a)
Compute the mean of X as follows:
(b)
Compute the variance of X as follows:
(c)
Compute the value of P(X < 3.7) as follows:
![P(X < 3.7)=\int\limits^{3.7}_{1.3}{\frac{1}{6.2-1.3}}\, dx\\\\=\frac{1}{4.9}\times [x]^{3.7}_{1.3}\\\\=\frac{3.7-1.3}{4.9}\\\\\approx 0.4898](https://tex.z-dn.net/?f=P%28X%20%3C%203.7%29%3D%5Cint%5Climits%5E%7B3.7%7D_%7B1.3%7D%7B%5Cfrac%7B1%7D%7B6.2-1.3%7D%7D%5C%2C%20dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B4.9%7D%5Ctimes%20%5Bx%5D%5E%7B3.7%7D_%7B1.3%7D%5C%5C%5C%5C%3D%5Cfrac%7B3.7-1.3%7D%7B4.9%7D%5C%5C%5C%5C%5Capprox%200.4898)
Thus, the value of P(X < 3.7) is 0.4898.
Answer:
this is the answer unless you did it backwards
Step-by-step explanation:
3.467406380027739e-4
You can simplifying 24/3 by dividing 24 by 3.
24÷3=8
The answer is 8.
Answer:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0
Is this simultaneous equations?
if yes then the first step is
make one of the unknown as the subject
