Answer:
The length of BC is needed because it is the side opposite ∠A.
Step-by-step explanation:
Given the right angles triangle as shown in the attachment, we can get sin(A) without using Pythagoras theorem. Instead we will use SOH CAH TOA trigonometry identity.
According to SOH:
Sin(A) = Opposite/Hypotenuse
Sin(A) = |BC|/|AB|
Opposite side of the triangle is the side facing ∠A.
Based on the formula, we will need to get the opposite side of the triangle which is length BC for us to be able to determine sinA since the hypotenuse is given.
we know that
The angle bisector, is the line or line segment that divides the angle into two equal parts.
So
in this problem
m∠ZXW=m∠WXY+m∠ZXY
m∠WXY=m∠ZXY
Part a) <u>Find the value of m∠WXY</u>
m∠ZXY=37°-----> given value
so
m∠WXY=m∠ZXY=37°
therefore
<u>the answer part a)</u>
The measure of angle WXY is equal to 37°
Part b) <u>Find the value of m∠ZXW</u>
m∠ZXW=m∠WXY+m∠ZXY
m∠ZXW=37°+37°=74°
therefore
<u>the answer part b) is</u>
the measure of angle ZXW is 74 degrees
Answer:
0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
![f(x) = \mu e^{-\mu x}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cmu%20e%5E%7B-%5Cmu%20x%7D)
In which
is the decay parameter.
The probability that x is lower or equal to a is given by:
![P(X \leq x) = \int\limits^a_0 {f(x)} \, dx](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%20%5Cint%5Climits%5Ea_0%20%7Bf%28x%29%7D%20%5C%2C%20dx)
Which has the following solution:
![P(X \leq x) = 1 - e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%201%20-%20e%5E%7B-%5Cmu%20x%7D)
The probability of finding a value higher than x is:
![P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%3E%20x%29%20%3D%201%20-%20P%28X%20%5Cleq%20x%29%20%3D%201%20-%20%281%20-%20e%5E%7B-%5Cmu%20x%7D%29%20%3D%20e%5E%7B-%5Cmu%20x%7D)
Mean of 4 minutes
This means that ![m = 4, \mu = \frac{1}{4} = 0.25](https://tex.z-dn.net/?f=m%20%3D%204%2C%20%5Cmu%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%3D%200.25)
Find the probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot:
This is:
![P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2)](https://tex.z-dn.net/?f=P%282%20%5Cleq%20X%20%5Cleq%20132%29%20%3D%20P%28X%20%5Cleq%20132%29%20-%20P%28X%20%5Cleq%202%29)
In which
![P(X \leq 132) = 1 - e^{-0.25*132} = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20132%29%20%3D%201%20-%20e%5E%7B-0.25%2A132%7D%20%3D%201)
![P(X \leq 2) = 1 - e^{-0.25*2} = 0.393469](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%3D%201%20-%20e%5E%7B-0.25%2A2%7D%20%3D%200.393469)
![P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2) = 1 - 0.393469 = 0.606531](https://tex.z-dn.net/?f=P%282%20%5Cleq%20X%20%5Cleq%20132%29%20%3D%20P%28X%20%5Cleq%20132%29%20-%20P%28X%20%5Cleq%202%29%20%3D%201%20-%200.393469%20%3D%200.606531)
0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.
Answer:
50, 40, 30, 250, 350
Step-by-step explanation:
1/2 = 0.5, 0.5 x 100 = <u>50</u> (0.5 -> 5 -> 50)
2/5 = 0.4 (10 / 5 [the denominator] = 2, 0.2 x 2 [the numerator] = 0.4), 0.4 x 100 = <u>40</u> (0.4 -> 4 -> 40)
3/10 = (10 / 10 [the denominator] = 1, 0.1 x 3 [the numerator] = 0.3), 0.3 x 100 = <u>30</u> (0.3 -> 3 -> 30)
5/2 = 2.5 (2 1/2), 2.5 x 100 = <u>250</u> (2.5 -> 25 -> 250)
7/2 = 3.5 (3 1/2), 3.5 x 100 = <u>350</u> (3.5 -> 35 -> 350)
Note: I'm not sure if I understand the question completely, but I changed the fraction into a decimal and multiplied it by 100. Not sure what it means by "<u><em>Divide</em></u><em> fraction</em>".