A men and a women decided to meet at a certain location. If each of them independently arrives at a time uniformly between 12pm
and 1pm, find the probability that the first to arrive has to wait longer than 10 mins?
1 answer:
Answer:
the probability that the first to arrive has to wait longer than 10 mins is 35/36.
Step-by-step explanation:
First you need to denote by X and Y the time past 12 noon that the man and woman arrive.
Then you have to compute P(X+10<Y)+P(Y+10<X), which by symmetry equals 2P(X+10<Y).
So basically you have to compute P(X+10<Y).
check the attachment for the rest of the step by step.
Answer should be 25/72
But since we want 2× of the above probability 2× 25/72 = 25/36 final answer.
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Answer:
Perimeter = 18.7 units
Area = 13.5 units²
Step-by-step explanation:
Perimeter of ADEC = AD + DE + EC + AC
Length of AD = 3 units
By applying Pythagoras theorem in ΔDBE,
DE² = DB² + BE²
DE² = 3² + 3²
DE = √18
DE = 4.24 units
Length of EC = 3 units
By applying Pythagoras theorem in ΔABC,
AC² = AB² + BC²
AC² = 6² + 6²
AC = √72
AC = 8.49 units
Perimeter of ADEC = 3 + 4.24 + 3 + 8.49
= 18.73 units
≈ 18.7 units
Area of ADEC = Area of ΔABC - Area of ΔBDE
Area of ΔABC =
=
= 18 units²
Area of ΔBDE =
=
= 4.5 units²
Area of ADEC = 18 - 4.5
= 13.5 units²
