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If the roads leading to an arena can together handle 3,500 vehicles/hour of arena-only traffic, at what time should the roads be

Svet_ta [12734]

The given road capacity is 3,500 vehicles per hour, and the expected number of vehicles arriving is 14,000. To calculate the time required for these vehicles to access the arena:
14,000 vehicles divided by 3,500 vehicles per hour equals 4 hours.

If the event is scheduled to commence at 7 p.m., the roads should close at: 7 p.m. minus 4 hours, which results in 3 p.m.

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2 months ago

3.7.6 .WP A state runs a lottery in which six numbers are randomly selected from 40 without replacement. A player chooses six nu

PIT_PIT [12445]

Answer:

a) 2.60x10^-7

b) 5.31x10^-5

c) 2.19x10^-3

Step-by-step explanation:

X=the number of hits

Probability can be determined by taking the number of successful outcomes and dividing it by the total number of possible outcomes.

Now

a) P(X=6)=P({1, 1, 1, 1, 1, 1})=6/40*5/39*4/38*3/37*2/36*1/35=2.60x10^-7

b) P(X=5)=P({0, 1, 1, 1, 1, 1})+P({1, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 1, 0}), and each scenario carries the same probability

P(X=5)=6P({1, 1, 1, 1, 1, 0})=6*(6/40*5/39*4/38*3/37*2/36*34/35)=5.31x10^-5

c) P(X=4)=P({0, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 0, 0}), where each case has an identical probability.

$P(X=4)=(^6_4)P({1, 1, 1, 1, 0, 0})=\frac{6!}{4!(6-4)!}(\frac{6}{40} \frac{5}{39} \frac{4}{38} \frac{3}{37} \frac{34}{36} \frac{33}{35} )=2.19\times 10^{-3}$

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2 months ago