You first separate the triangle from the rectangle. Then multiply one side be one side.
List all the different combinations with total cost:
Airplane + van = 350 + 60 = $410
Airplane + cab = 350 + 40 = $390
Bus + van = 150 + 60 = $210
Bus + cab = 150 + 40 = $190
Train + van = 225 + 60 = $285
Train + cab = 225 + 40 = $265
There are 6 different combinations.
There are 2 that are more than $300
The probability would be quantity higher than 300 / total combinations.
Probability = 2/6 = 1/3
Update the question so it's on-topic for Mathematics Stack Exchange. Closed 4 years ago. How many words can be made by using the letters of the word ENGLAND such that words begin and end with a vowel? I couldn't understand how come the answer to this question is 120
we know the segment QP is an angle bisector, namely it divides ∡SQR into two equal angles, thus ∡1 = ∡2, and ∡SQR = ∡1 + ∡2.
![\bf \begin{cases} \measuredangle SQR = \measuredangle 1 + \measuredangle 2\\\\ \measuredangle 2 = \measuredangle 1 = 5x-7 \end{cases}\qquad \qquad \stackrel{\measuredangle SQR}{7x+13} = (\stackrel{\measuredangle 1}{5x-7})+(\stackrel{\measuredangle 2}{5x-7}) \\\\\\ 7x+13 = 10x-14\implies 13=3x-14\implies 27=3x \\\\\\ \cfrac{27}{3}=x\implies 9=x \\\\[-0.35em] ~\dotfill\\\\ \measuredangle SQR = 7(9)+13\implies \measuredangle SQR = 76](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20%5Cmeasuredangle%20SQR%20%3D%20%5Cmeasuredangle%201%20%2B%20%5Cmeasuredangle%202%5C%5C%5C%5C%20%5Cmeasuredangle%202%20%3D%20%5Cmeasuredangle%201%20%3D%205x-7%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cmeasuredangle%20SQR%7D%7B7x%2B13%7D%20%3D%20%28%5Cstackrel%7B%5Cmeasuredangle%201%7D%7B5x-7%7D%29%2B%28%5Cstackrel%7B%5Cmeasuredangle%202%7D%7B5x-7%7D%29%20%5C%5C%5C%5C%5C%5C%207x%2B13%20%3D%2010x-14%5Cimplies%2013%3D3x-14%5Cimplies%2027%3D3x%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B27%7D%7B3%7D%3Dx%5Cimplies%209%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cmeasuredangle%20SQR%20%3D%207%289%29%2B13%5Cimplies%20%5Cmeasuredangle%20SQR%20%3D%2076)
Answer:
Step-by-step explanation:
The segment joining an original point with its rotated image forms a chord of the circle of rotation containing those two points. The center of the circle is the center of rotation.
This means you can find the center of rotation by considering the perpendicular bisectors of the segments joining points with their images. Here, the only proposed center that is anywhere near the perpendicular bisector of DE is point M.
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Segment AD is perpendicular to corresponding segment FE, so the angle of rotation is 90°. (We don't know which way (CW or CCW) unless we make an assumption about which is the original figure.)
