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3 years ago

Help me with this plz

Mathematics

1 answer:

andriy [413]3 years ago

7 0

1. angle X and and angle 80 are same side interior angles
2. congruent

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What is the value of -36+(-4/9)

zalisa [80]

$- 36 \times 9 = - 324$
-324/9 -4/9
-328/9

5 0

4 years ago

A large clock in a town square has a minute hand that is 10 feet long. Which value is the best estimate of

nataly862011 [7]

Answer: 240 degrees

Step-by-step explanation: minute hand makes 360 degree circle in hour.

There is no influence for the length of hand in this case.

40 minutes is 2/3 hours. 2/3 ·360 degrees

8 0

3 years ago

Pleas explain how you get 4 5/24 from 6 1/8-1 11/12? I got 4 2/24 but after using my calculator, I got 5/24.

Savatey [412]

$6 \frac{1}{8}-1 \frac{11}{12}= \frac{49}{8}- \frac{23}{12}= \frac{49*3-23*2}{24} = \frac{147-46}{24}= \frac{101}{24}=4 \frac{5}{24}$

Ноpe this helps

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)

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

Thus,

$P(Y>X) = 1\cdot \frac{1}{4} + \frac{3}{8}\cdot \frac{3}{4} = \frac{17}{32}$

5 0

4 years ago

Pls help with this math problem

valentinak56 [21]

-1.5. -1 1/2. -3/2

yeah so its this i’m 99.98% sure

7 0

2 years ago