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Andrew [12]
3 years ago
15

A traffic engineer records a sample of the types of vehicles that cross a bridge. She counts 112 passenger cars, 18 light trucks

, and 20 heavy trucks.
What is the APPROXIMATE probability that a random vehicle crossing the bridge is a passenger car?
Mathematics
1 answer:
Bond [772]3 years ago
8 0

Answer:

<em>The approximate probability that a random vehicle is a passenger car is 74.67%</em>

Step-by-step explanation:

<u>Probability</u>

The traffic engineer records 112 passenger cars, 18 light trucks, and 20 heavy trucks. The total of vehicles recorded is n=112+18+20 = 150.

We are required to find the probability that a randomly selected vehicle crossing the bridge is a passenger car. There are 112 passenger cars, thus the probability is:

\displaystyle p=\frac{112}{150}

p = 0.7467

p = 74.67%

The approximate probability that a random vehicle is a passenger car is 74.67%

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Step-by-step explanation:

40/8 = 5 dogs per hour

12 x 5 = 60 dogs in 12 hours

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Y=-x^2-x-3 find the X intercept for The parabola define by this equation
KengaRu [80]
To find the X intercept make y =0. Then factor out a negative on the side with the equation for each of the terms, so it becomes
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7 0
3 years ago
Write an expression meaning 4 less than 5 times a number
Rudiy27

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4 0
3 years ago
Read 2 more answers
From an industrial area 70 companies were selected at random and 45 of them were panning for expansion next year. Find 95% confi
Lana71 [14]

Answer:

Confidence limit = [52.8%, 75.2%]

Step-by-step explanation:

P=\frac{45}{70}= 0.64

(1-P)=1-0.64=0.36

n= 70

P ± z \sqrt{\frac{P(1-P)}{n} }

where the value z will be taken from the z-table for 95% confidence interval

1-0.95= 0.05/2= 0.025

0.95+0.025= 0.0975

From the z-table the value of z corresponding to 0.0975 is 1.96

0.64 ± 1.96 \sqrt{\frac{0.64*0.36}{70} }

0.64 ± 1.96 (0.057)

0.64 ± 0.112

64%% ± 11.2%

so the confidence interval is

64+11.2=75.2%

64-11.2=52.8%

[52.8, 75.2]

8 0
3 years ago
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