We can construct a unique equilateral triangle if we know its___________.
1 answer:
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Answer:
0.1333 = 13.33% probability that bridge B was used.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Arrives home by 6 pm
Event B: Bridge B used.
Probability of arriving home by 6 pm:
75% of 1/3(Bridge A)
60% of 1/6(Bridge B)
80% of 1/2(Bridge C)
So
Probability of arriving home by 6 pm using Bridge B:
60% of 1/6. So
Find the probability that bridge B was used.
0.1333 = 13.33% probability that bridge B was used.
Check the picture below.
now, you can pretty much count the units off the grid for the segments ST and RU, so each is 7 units long, and are parallel, meaning that the other two segments are also parallel, and therefore the same length each.
so we can just find the length for hmmmm say SR, since SR = TU, TU is the same length,
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ S(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad R(\stackrel{x_2}{-5}~,~\stackrel{y_2}{5})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ SR=\sqrt{[-5-(-2)]^2+[5-1]^2}\implies SR=\sqrt{(-5+2)^2+(5-1)^2} \\\\\\ SR=\sqrt{(-3)^2+4^2}\implies SR=\sqrt{25}\implies SR=5](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AS%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20%0AR%28%5Cstackrel%7Bx_2%7D%7B-5%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ASR%3D%5Csqrt%7B%5B-5-%28-2%29%5D%5E2%2B%5B5-1%5D%5E2%7D%5Cimplies%20SR%3D%5Csqrt%7B%28-5%2B2%29%5E2%2B%285-1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ASR%3D%5Csqrt%7B%28-3%29%5E2%2B4%5E2%7D%5Cimplies%20SR%3D%5Csqrt%7B25%7D%5Cimplies%20SR%3D5)
sum all segments up, and that's perimeter.
now, you can pretty much count the units off the grid for the segments ST and RU, so each is 7 units long, and are parallel, meaning that the other two segments are also parallel, and therefore the same length each.
so we can just find the length for hmmmm say SR, since SR = TU, TU is the same length,
sum all segments up, and that's perimeter.