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Alinara [238K]
3 years ago
11

The graph shows the cost of some taxi journeys. work out a formula for C in terms of n.

Mathematics
1 answer:
Serga [27]3 years ago
7 0

Given:

The graph of cost of taxi with respect to number of miles.

To find:

The formula for C in terms of n.

Solution:

In the given graph x axis represents the number of miles (n) and y-axis represents the cost of taxi (C).

From the given graph it is clear that, the line passes through the points (0,3) and (5,6). So, the equation of line is

y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

y-3=\dfrac{6-3}{5-0}(x-0)

y-3=\dfrac{3}{5}(x)

Adding 3 on both sides, we get

y=\dfrac{3}{5}x+3

Putting y=C and x=n, we get

C=\dfrac{3}{5}n+3

Therefore, the required formula is C=\dfrac{3}{5}n+3.

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Answer: The number 12 represents admission charge in Mariana's equation.

Step-by-step explanation:

As per given ,

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c = 1.50x + 12                                       (ii)

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Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Us
grigory [225]

Answer:

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

Step-by-step explanation:

Previous concepts

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".  

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".

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

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Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

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Question 3
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Answer:

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Step-by-step explanation:

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