HELP ME I am not really sure how to explain this !!!

Simplify 5x3 + 5(6x3 - 564).

Thepotemich [5.8K]

I would download photocalc

8 0

3 years ago

Point A is located at (3,-6) and point A is located at (1-2). What is the scale factor?

Anuta_ua [19.1K]

Answer:

The scale factor is 3.

Step-by-step explanation:

Let the origin O(0,0) is the reference point and the coordinates of the point A are given by (3,-6).

Therefore, the distance OA is given by $\sqrt{(3 - 0)^{2} + (- 6 - 0)^{2}} = 3\sqrt{5}$ units.

Again, the coordinates of point A' are (1,-2).

Therefore, the distance OA' is given by $\sqrt{(1 - 0)^{2} + (- 2 - 0)^{2}} = \sqrt{5}$ units.

Hence, the scale factor is $\frac{OA}{OA'} = \frac{3\sqrt{5} }{\sqrt{5}} = 3$ . (Answer)

4 0

3 years ago

Let X equal the number of hours that a randomly selected person from City A reads the news online each day. Suppose that X has a

Nina [5.8K]

Answer:

$P(Y>X) = \frac{17}{32}$

Step-by-step explanation:

Recall that since X is uniformly distributed over the set [1,4] we have that the pdf of X is given by $f(x) = \frac{1}{3}$ if $1\leq x \leq 4$ and 0 otherwise. In the same manner, the pdf of Y is given by $g(y) = \frac{1}{4}$ if $1\leq y\leq 5$ and 0 otherwise.

Note that if Y is in the interval (4,5] then Y>X by default. So, in this case we have that P(Y>X| y in (4,5]) = 1. We want to calculate the probability of having Y in that interval . That is

$P(Y>4) = \int_{4}^{5}\frac{1}{4}dy = \frac{1}{4}$ . Thus, $P(Y\leq 4 ) = 1 - P(Y>4)= \frac{3}{4}$ .

We want to proceed as follows. Using the total probability theorem, given two events A, B we have that

$P(A) = P(A|B)\cdot P(B) + P(A|B^c) \cdot P(B^c)$ In this case, A is the event that Y>X and B is the event that Y is in the interval (4,5].

If we assume that X and Y are independent, then we have that the joint pdf of X,Y is given by $h(x,y) = f(x)g(y) = \frac{1}{12}$ when $1\leq x \leq 4, 1\leq y \leq 4$ . We can draw the region were Y>X and the function h(x,y) is different from 0. (The drawing is attached). This region is described as follows: $1\leq x \leq 4$ and $x\leq y \leq 4$ , then (the specifics of the calculations of the integrals are ommitted)

$P(Y>X| y \in [1,4)) = \int_{1}^{4}\int_{x}^{4}\frac{1}{12}dy dx = \frac{1}{12}\int_{1}^{4} (4-x) dx = \frac{1}{12}\left.(4x-\frac{x}{2})\right|_{1}^{4}= \frac{1}{12}(4\cdot 4 - \frac{4^2}{2}-(4-\frac{1}{2}) = \frac{9}{2\cdot 12} = \frac{3}{8}$