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Dovator [93]
3 years ago
13

The cycle time for trucks hauling concrete to a highway construction site is uniformly distributed over the interval 40 to 75 mi

nutes. What is the probability that the cycle time exceeds 50 minutes if it is known that the cycle time exceeds 45 minutes
Mathematics
1 answer:
jeka57 [31]3 years ago
8 0

Answer:

0.8333 = 83.33% probability that the cycle time exceeds 50 minutes if it is known that the cycle time exceeds 45 minutes

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value higher than x is given by:

P(X > x) = \frac{b - x}{b - a}

Uniformly distributed over the interval 40 to 75 minutes.

This means that a = 40, b = 75

It is known that the cycle time exceeds 45 minutes

This means that we can use a = 45

What is the probability that the cycle time exceeds 50 minutes?

P(X > 50) = \frac{75 - 50}{75 - 45} = 0.8333

0.8333 = 83.33% probability that the cycle time exceeds 50 minutes if it is known that the cycle time exceeds 45 minutes

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OlgaM077 [116]

so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.


\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.


\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

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