Can you guys please help. I really don't know how to solve this

Mathematics

1 answer:

Nonamiya [84]3 years ago

3 0

The inequalities that can be used to represent this problem are x < 0 or x > 5. On the number line, you will want to place an open circle on both 0 and 5. The open circle signifies that x is not equal to these values, but is greater or less than. On the open circle on 0, draw an arrow pointing left, toward the negative values. On the open circle on 5, draw an arrow pointing right, toward the positive values.

Hope this helps!! :)

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A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon

melamori03 [73]

Answer:

51/4

Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.

X~ Uniform(0,100)

Then the probability mass function is given as follows.

$f(x) = P(X=x) = 1/100 \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0 \,\,\,\, \text{otherwise}$

Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located 70 meters away from city A then the mid point between 20 and 70 is (70+20)/2 = 45 then we can represent D as follows

$D(x) =\left\{ \begin{array}{ll} x & \mbox{if } 0\leq x \leq 20 \\ x-20 & \mbox{if } 20\leq x < 45\\ 70-x & \mbox{if } 45 \leq x \leq 70\\ x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.$

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable $Y = D(X)$ , $Y$ is a random variable as well, remember that there is a theorem that says that

$E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx$

Where $f(x)$ is the probability mass function of X. Using the information of our problem

$E[Y] = \int\limits_{-\infty}^{\infty} D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx \bigg]\\= \frac{51}{4} = 12.75$