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kondaur [170]
3 years ago
14

Dana leaves Las Vegas for LA at 2 p.m. driving at 55 mph. At 4 p.m. Lance leaves LA for Las Vegas driving at 45 mph along the sa

me route. If the cities are 260 miles, what time do they meet?
Mathematics
1 answer:
NNADVOKAT [17]3 years ago
4 0

Answer: They meet after 1 hour 42 minutes.

Step-by-step explanation:

Since we have given that

Dana leaves Las Vegas for LA at 2 p.m. driving at 55 mph.

Let the time taken by Dana be 't'.

Distance traveled by Dana would be 55t.

At 4 p.m. Lance leaves LA for Las Vegas driving at 45 mph along the same route.

It means after 2 hours Lance leave for LA.

So, time taken by Lance be 't-2'.

Distance traveled by Lance would be 45(t-2)

Total distance  = 260 miles

According to question, it becomes,

55t+45(t-2)=260\\\\55t+45t-90=260\\\\100t=260-90\\\\100t=170\\\\t=1.7\ hours=1\dfrac{7}{10}=1\ hour\ and\ \dfrac{7\times 60}{10}\ minutes=1\ hour\ 42\ minutes

Hence, they meet after 1 hour 42 minutes.

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t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

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Step-by-step explanation:

Data given and notation  

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n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

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Alternative hypothesis:\mu \neq 1.35  

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t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

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df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

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Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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