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Tema [17]
4 years ago
12

A certain NYC taxi driver has decided to start charging a rate of r cents per person per mile. How much, in dollars, would it co

st 3 people to travel x miles if he decides to give them a 50% discount?
a. 3xr/2
b. 3x/200r
c. 3r/200x
d. 3xr/200
e. xr/600
Mathematics
1 answer:
katrin2010 [14]4 years ago
3 0

Answer:

\frac{3xr}{200}

Step-by-step explanation:

A certain NYC taxi driver has decided to start charging a rate of r cents per person per mile

rate is cents . convert it in to dollars

1 cent = 1/100 dollars

r cents =  r/100 dollars

Each people travels x miles

cost is \frac{r}{100} \cdot x

For 3 people the cost is

3\frac{r}{100} \cdot x

Discount is 50%, so we divide by 2

\frac{3\frac{r}{100} \cdot x}{2} =\frac{3xr}{200}

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a) 183-1.984\frac{20}{\sqrt{100}}=179.032    

183+1.984\frac{20}{\sqrt{100}}=186.968    

So on this case the 95% confidence interval would be given by (179.032;186.968)    

b) 190-1.984\frac{23}{\sqrt{100}}=185.437    

190+1.984\frac{23}{\sqrt{100}}=194.563    

So on this case the 95% confidence interval would be given by (185.437;194.563)    

c) For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Part a : Summer

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=100-1=99

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that t_{\alpha/2}=1.984

Now we have everything in order to replace into formula (1):

183-1.984\frac{20}{\sqrt{100}}=179.032    

183+1.984\frac{20}{\sqrt{100}}=186.968    

So on this case the 95% confidence interval would be given by (179.032;186.968)    

Part b: Winter

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=100-1=99

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that t_{\alpha/2}=1.984

Now we have everything in order to replace into formula (1):

190-1.984\frac{23}{\sqrt{100}}=185.437    

190+1.984\frac{23}{\sqrt{100}}=194.563    

So on this case the 95% confidence interval would be given by (185.437;194.563)    

Part c

For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

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