50 POINTS: The graph of a system of equations with the same slope will have no solutions. WITH EXPLINATION PLEASE

1 answer:

8 0

A system of two linear equations can be graphed as two lines.
The solution of the system of equations is a point that is part of both lines.
As a graph, you see two intersecting lines. The only point on both lines is the intersection of the lines. This is the solution of the system of equations.

Now think of two parallel lines. They never intersect. Parallel lines have the same slope. Since there is no point that is part of both lines, there is no solution to the system of equations.

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Determine whether the relation below is also a function.

Bess [88]

Answer:

B. Not a function

Step-by-step explanation:

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3 years ago

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

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