Answer:
a) P(X > 10) = 0.6473
b) P(X > 20) = 0.4190
c) P(X < 30) = 0.7288
d) x = 68.87
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:

In which
is the decay parameter.
The probability that x is lower or equal to a is given by:

Which has the following solution:

The probability of finding a value higher than x is:

Mean equal to 23.
This means that 
(a) P(X >10)

So
P(X > 10) = 0.6473
(b) P(X >20)

So
P(X > 20) = 0.4190
(c) P(X <30)

So
P(X < 30) = 0.7288
(d) Find the value of x such that P(X > x) = 0.05
So






Answer:

Step-by-step explanation:
![f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%7B2x%5E3-1%7D%3D4%5Cleft%282x%5E3-1%5Cright%29%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5Cf%27%28x%29%3D4%5Ccdot%5Cdfrac%7B1%7D%7B2%7D%282x%5E3-1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ccdot3%5Ccdot2x%5E2%3D%5Cdfrac%7B12x%5E2%7D%7B%282x%5E3-1%29%5E%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cdfrac%7B12x%5E2%7D%7B%5Csqrt%7B2x%5E3-1%7D%7D%5C%5C%5C%5C%5Ctext%7Bused%7D%5C%5C%5C%5C%5Csqrt%7Ba%7D%3Da%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5Cbigg%5Bf%5Cleft%28g%28x%29%5Cright%29%5Cbigg%5D%27%3Df%27%28g%28x%29%29%5Ccdot%20g%27%28x%29%5C%5C%5C%5C%5Cbigg%5Bnf%28x%29%5Cbigg%5D%27%3Dnf%27%28x%29%5C%5C%5C%5C%28x%5En%29%27%3Dnx%5E%7Bn-1%7D)
<span>he range if -5 ≤ y ≤ 5.
The period is 2π/(π/2) = 4.
The graph passes through (0, 0), has a maximum at (1, 5), crosses the x-axis at (2, 0), then has a minimum at (3. -5), and crosses the x-axis again at (4, 0). </span><span>
</span>
Answer:
y= -1x+1
Step-by-step explanation: