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?

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 probability2× 25/72 = 25/36 final answer.

answer is : muchas/tipicas

have a nice day!

~recklesschicken

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.