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Sauron [17]
3 years ago
14

A car is leaving New York. After 30 minutes, the car is 25 miles from New York, and after 3 hours the car is 200 miles from New

York. What is the car's average speed (miles per hour) between the 30 minute and 3 hour time frame
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

57.1428571429 miles per hour <em>Round off the answer to what your teacher asked.</em>

Step-by-step explanation:

Since we want final answer in miles per hour, lets change the 30 mins to 0.5 hours

Phase 1 =25miles per 0.5 hours

Phase 2 = (200miles-25miles) per 3 hours

Speed = distance/time

Average speed = total distance/ time

Total distance is 200miles because both phases say from new york that means it travels 25 miles away from new york then 3 hours later it is 200 miles away from new york

total time =3 hours+0.5 hours = 3.5 hours

Average speed = 200/3.5

Average speed = 57.1428571429 miles per hour

<em>Rate brainliest please</em>

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Answer:

B

Step-by-step explanation:

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3 years ago
Eighty percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the air
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Answer:

a) P(B'|A) = 0.042

b) P(B|A') = 0.625

Step-by-step explanation:

Given that:

80% of the light aircraft that disappear while in flight in a certain country are subsequently discovered

Of the aircraft that are discovered, 63% have an emergency locator,

whereas 89% of the aircraft not discovered do not have such a locator.

From the given information; it is suitable we define the events in order to calculate the probabilities.

So, Let :

A = Locator

B = Discovered

A' = No Locator

B' = No Discovered

So; P(B) = 0.8

P(B') = 1 - P(B)

P(B') = 1- 0.8

P(B') = 0.2

P(A|B) = 0.63

P(A'|B) = 1 - P(A|B)

P(A'|B) = 1- 0.63

P(A'|B) = 0.37

P(A'|B') = 0.89

P(A|B') = 1 - P(A'|B')

P(A|B') = 1 - 0.89

P(A|B') = 0.11

Also;

P(B ∩ A) = P(A|B) P(B)

P(B ∩ A) = 0.63 × 0.8

P(B ∩ A) = 0.504

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P(B ∩ A') = 0.37 × 0.8

P(B ∩ A') = 0.296

P(B' ∩ A) = P(A|B') P(B')

P(B' ∩ A) = 0.11 × 0.2

P(B' ∩ A) = 0.022

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P(B' ∩ A') = 0.89 × 0.2

P(B' ∩ A') = 0.178

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P(A) = 0.504 + 0.022

P(A) = 0.526

P(A') = 1 - P(A)

P(A') = 1 - 0.526

P(A') =  0.474

The probability that it will not be discovered given that it has an emergency locator is,

P(B'|A) =  P(B' ∩ A)/P(A)

P(B'|A) = 0.022/0.526

P(B'|A) = 0.042

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The probability that it will be discovered given that it does not have an emergency locator is:

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