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Fiesta28 [93]
1 year ago
14

Yuto left his house at 10 a.m. to go for a bike ride. By the time Yuto’s sister Riko left their house, Yuto was already 5.25 mil

es along the path they both took. If Yuto’s average speed was 0.25 miles per minute and Riko’s average speed was 0.35 miles per minute, over what time period in minutes, t, starting from when Riko left the house, will Riko be behind her brother?
Mathematics
2 answers:
Nadusha1986 [10]1 year ago
7 0

Answer:

The solution means that Riko will be behind Yuto from the time she leaves the house, which corresponds to t = 0, until the time she catches up to Yuko after 52.5 minutes, which corresponds to t = 52.5. The reason that t cannot be less than zero is because it represents time, and time cannot be negative.

Hope this helps!!! :) Have a great day/night.

Akimi4 [234]1 year ago
4 0

Answer:

52 minutes and 30 seconds

Step-by-step explanation:

You know that Yuto has ridden for 5.25 miles when Riko left their house and you need to know and what time they will be together:

Then you can say that:

5.25 miles+(yutos speed)*t= (Rikos speed)*t

when t=Time in minutes when they will be together

5.25 miles+(0.25miles/min)*t= (0,35miles/min)*t

5.25miles=(0.35miles/min-0.25miles/min)*t

5.25miles/(0.35miles/min-0.25miles/min)=t

t=52.5 min =52 minutes and 30 seconds

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71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

Binomial probability distribution

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.2, n = 200. So

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0.8554 - 0.1446 = 0.7108

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