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Scilla [17]
3 years ago
10

A person takes a trip, driving with a constant peed of 89.5 km/h, except for a 22.0-min rest stop. If the peron's average speed

is 77.8 km/h, (a) how much time is spent on the trip and (b) how far does the person travel?
Mathematics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

a) The person traveled 2.83 hours.

b) The person travels 220.17 kilometers.

Step-by-step explanation:

We have that the speed is the distance divided by the time. Mathematically, that is

s = \frac{d}{t}

(a) how much time is spent on the trip and

The peron's average speed is 77.8 km/h, which means that s = 77.8

The person distance traveled is:

22 min is 22/60 = 0.37h.

So for  the time t1, the person traveled at a speed of 89.5 km/h. Which has a distance of 89.5*t1.

For 0.37h, the person was at a stop, so she did not travel. This means that the total distance is

d = 89.5t1 + 0 = 89.5t1

The total time is the time traveling t and the stoppage time 0.37. So

t = t1 + 0.37

We want to find t1, which is the time that the person was driving.

So

s = \frac{d}{t}

77.8 = \frac{89.5t1}{t1 + 0.37}

77.8t1 + 77.8*0.37 = 89.5t1

11.7t1 = 28.786

t1 = \frac{28.786}{11.7}

t1 = 2.46

The total time is

t = t1 + 0.37 = 2.46 + 0.37 = 2.83

The person traveled for 2.83 hours.

(b) how far does the person travel?

The person traveled 2.46 hours at an average speed of 77.8 km/h. So

s = \frac{d}{t}

77.8 = \frac{d}{2.83}

d = 77.8*2.83 = 220.17

The person travels 220.17 kilometers.

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3 years ago
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Answer:

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Now we can find the p value. Since we have a bilateral test the p value would be:  

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Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

Do Not reject H0

Step-by-step explanation:

Information provided

n=200 represent the sample size slected

X=155 represent the cell phone owners used text messaging

\hat p=\frac{155}{200}=0.775 estimated proportion of cell phone owners used text messaging

p_o=0.73 is the value to verify

\alpha=0.1 represent the significance level

We need to conduct a z test for a proportion

z would represent the statistic

p_v represent the p value

System of hypothesis

We want to verify if the true proportion of cell phone owners used text messaging is different from 0.73 so then the system of hypothesis are:

Null hypothesis:p=0.73  

Alternative hypothesis:p \neq 0.73  

The statistic to check this hypothesis is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the data given we got:

z=\frac{0.775 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=1.433  

Now we can find the p value. Since we have a bilateral test the p value would be:  

p_v =2*P(z>1.433)=0.152  

Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

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Step-by-step explanation:

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