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geniusboy [140]
3 years ago
12

Two cars leave Phoenix and travel along roads 90 degrees apart. If Car 1 leaves 60 minutes earlier than Car 2 and averages 40 mp

h and if Car 2 averages 50 mph, how far apart will they be after Car 1 has traveled 3.5 hours?
Mathematics
1 answer:
prohojiy [21]3 years ago
5 0

Answer:

The two cars will be almost 188 miles far from each other.

Step-by-step explanation:

Travel Time for Car 1 = t = 3.5 hours

Travel time for Car 2 = t-1 = 3.5 - 1 = 2.5 hours

Average speed of car 1 = 40 mph

Average speed of car 2 = 50 mph

Distance traveled by Car 1 = 40*3.5 = 140 miles

Distance Traveled by Car 2 = 50*2.5 = 125 miles

As both the roads are at a 90 degree angle. The path of the two cars and the joining line of their final position forms a right angle triangle where:

altitude = a = 140

base = b = 125

Distance of cars after 3.5 hours = c = ?

According to Pythagoras theorem:

c^2 = a^2 + b^2

c^2 = 140² + 125²

c² = 19600+15625

c = √35225

c = 187.68

Almost 188 miles.

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Answer:

The standard error of the mean is 0.0783.

Step-by-step explanation:

The Central Limit Theorem helps us find the standard error of the mean:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

The standard deviation of the sample is the same as the standard error of the mean. So

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In this problem, we have that:

\sigma = 0.35, n = 20

So

SE_{M} = \frac{\sigma}{\sqrt{n}}

SE_{M} = \frac{0.35}{\sqrt{20}}

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