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Ronch [10]
3 years ago
11

A long distance runner starts at the beginning of a trail and runs at a rate of 6 miles per hour. One hour later, a cyclist star

ts at the beginning of the trail and travels at a rate of 14 miles per hour. What is the amount of time that the cyclist travels before overtaking the runner?
Mathematics
1 answer:
sasho [114]3 years ago
5 0
The answer to this is 50 minutes. by then the runner has gone 11 miles in an hour and 50min and the cyclist has gone 11 miles also 
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Ryan earns x dollars every seven days. Write an expression for how much Ryan earns in one day. Ryan’s spouse Lee is paid twice a
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Answer:

Step-by-step explanation:

R=x/7 let's say R is what Ryan earned, and so he earns x dollars every seven days

L=2(R)

L=2(x/7)Let's say L is what Lee earned, she earned twice as much as Ryan so 2(x/7)

Total=2(x/7)+x/7

Total=3(x/7)   the total earnings is the sum of x/7 and 2x/7 is 3x/7

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The function is used to convert temperature from Celsius (C) to Fahrenheit (F). The function C(K) = K − 273.15 is used to conver
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Answer:

F(k)= \frac{9}{5}(k - 273.15) + 32

 

Step-by-step explanation:

Given

F(c)= \frac{9}{5}c + 32

C(k) = k - 273.15

Required

Function from K to F

This implies that, we calculate F(k)

Given that:

F(c)= \frac{9}{5}c + 32

Substitute c(k) for c

F(c(k))= \frac{9}{5}c(k) + 32

Substitute: C(k) = k - 273.15

F(c(k))= \frac{9}{5}(k - 273.15) + 32

Hence:

F(k)= \frac{9}{5}(k - 273.15) + 32

8 0
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Explain the difference between the two inequalities. how does this affect your method of solution?
AfilCa [17]
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3 years ago
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Answer:

P (y < 6) = \frac{P(y- \mu)}{\sigma}  \mu}{\sigma} \\\\=P(z

Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618

Step-by-step explanation:

Normal with mean 6.88 minutes and

a standard deviation of 0.57 minutes.

Choose a student at random from this group and call his time for the mile Y. Find P(Y<6)

\mu = 6.88

\sigma = 0.57

y ≈ normal (μ, σ)

The z score is the value decreased by the mean divided by the standard deviation

P (y < 6) = \frac{P(y- \mu)}{\sigma}  \mu}{\sigma} \\\\=P(z

Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618

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Answer:

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Step-by-step explanation:

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