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Geometric Probablity...please help me!?

k well i really need help on this probelm...please help me..or at least teach me how to do it or how to start on it...

the traffic light at fourth and Commercial Streets repeat themselves in 60-second cycles. Ms. Li regulary has students drive on Fourth street through Commercial Street intersections. By experience , she knows that they will face a red light 60% of the time . Use this information to estimate how long the Fourth Street light is red during each 1-min cycle.

please i really dont understand it..could you please help me..thanks!

it is a 60 second cycle (1 minute)
the light is red 60% of the time
60% of 60 is .60 * 60 = 36 minutes

so the answer is 36 minutes

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Math Story Poblem, I am not very good at these and need help with a starting point, please!?

Here is the problem: A modernistic painting consists of triangles, rectangles & pentagons, all drawn so as to not overlap or share sides. Within each rectangle are drawn 2 red roses & each pentagon contains 5 carnations. How many triangles, rectangles & pentagons appear in the painting if the painting contains a total of 40 geometric figures,153 sides of geometric figures and 72 flowers?

How do I set this up so that I can solve it? I am sure if I can get it set up that I can solve it! Can you help me?

Thanks in advance,

Jennifer

That is a ridiculous problem, but this is how you do it. Set up a system. Let x be the triangles, y be the rectangles, and z be the pentagons.

So, now you start setting up equations to use what you know. 3x+4y+5z=153 (this models the sides of the figures)
x+y+z=40 (modelling that there are 40 shapes together)
2y+5z= 72 models the number of flowers. there is no x term, because there was no info given concerning the flowers in a triangle.

So, to solve the system, use graphing, or substitution - whatever method you know best....

Here is using substitution:

x=40-y-z
3(40-y-z)+4y+5z=153
Distribute: 120-3y-3z+4y+5z=153
Simp: y+2z=33
y=33-2z

Put that into the equation that only has y and z.....

2(33-2z)+5z=72
z=6

So, x+y+6=40
x+y=34
x=34-y

Sub that in to the first eq, and you end up with z=6, y=21and x=13.....

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