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katrin [286]
3 years ago
9

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.63 m/s. Four seconds later,

the bicyclist hops on his bike and accelerates at 2.11 m/s2 until he catches his friend. How much time does it take until he catches his friend
Mathematics
1 answer:
castortr0y [4]3 years ago
5 0

From the moment the friend passes the bicyclist, his friend covers a distance over time t of (3.63 m/s)*t.

The bicyclist covers a distance of 1/2*(2.11 m/s^2)*t^2. They meet when these distances are equal:

3.63 t = 1.055 t^2  ==>  1.055 t^2 - 3.63 t = 0

==>  t = 0 s   or   t = 3.44 s

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Don't repost because somebody will answer (probably)

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3 years ago
A paper describes a study in which researchers observed wait times in coffee shops in Boston. Both wait time and gender of the c
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Answer:

t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65  

p_v =2*P(t_{(284)}

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

So the p value is a very low value and using any significance level giveen \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the means are different.

Step-by-step explanation:

1) Data given and notation

\bar X_{M}=85.8 represent the mean for the males

\bar X_{F}=113.4 represent the mean for females

s_{M}=50 represent the sample standard deviation for males

s_{F}=75 represent the sample standard deviation for female

n_{M}=145 sample size for the group poisoned

n_{F}=141 sample size for the group unpoisoned

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for the two groups are equal or not , the system of hypothesis would be:

Null hypothesis:\mu_{M} = \mu_{F}

Alternative hypothesis:\mu_{M} \neq \mu_{F}

If we analyze the size for the samples both are higher than 30 but the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{M}-\bar X_{F}}{\sqrt{\frac{s^2_{M}}{n_{M}}+\frac{s^2_{F}}{n_{F}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We can replace in formula (1) the results obtained like this:

t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65  

4) Statistical decision

The first step is calculate the degrees of freedom, on this case:

df=n_{M}+n_{F}-2=145+141-2=284

Since is a two tailed test the p value would be:

p_v =2*P(t_{(284)}

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

So the p value is a very low value and using any significance level giveen \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the means are different.

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Step-by-step explanation:

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