This is a uniform distribution with parameters a = 1.00 and b =5.00. With these conditions, the following probability distribution functions can be applied:

$P(X \leq x)=\frac{x-a}{b-a}=\frac{x-1.00}{5.00-1.00}\\P(X \leq x) = \frac{x-1}{4}\\P(X \geq x) = 1- P(X \leq x) \\P(X \geq x)=1- \frac{x-1}{4}$

Therefore, the probability that X ≥ 4:00 p.m. is given by

$P(X \geq 4)=1- \frac{4-1}{4}\\P(X \geq 4)=0.25=25\%$

Thus, 25% of deliveries are made after 4:00 p/m.

7 0

2 years ago

Which of the following are solutions to | x-1|= 5x + 2 ? Check all that apply.

agasfer [191]

By definition of absolute value, if $x-1\ge0$ , then $|x-1|=x-1$ , so

$|x-1|=5x+2\implies x-1=5x+2\implies4x=-3\implies x=-\dfrac34$

On the other hand, if $x-1$ , then $|x-1|=-(x-1)=1-x$ , and

$|x-1|=5x+2\implies1-x=5x+2\implies6x=-1\implies x=-\dfrac16$

4 0

2 years ago