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3 years ago

It cost $75 to rent a limo, plus $20 per hr. Write an equation to represent the total cost,c, of renting the limo for T hours.

1 answer:

5 0

We must create the equation from the information. Well, we know that there is a 75 dollar fee right off the top, and there is the word "plus" in the information meaning there will be adding.

1. Add the given cost plus the addition sign.

75 + ___ = ?

Now, we must find out what fills in "__." It says that it costs $20 per hour to rent the limo. However, it doesnt tell you how many hours it is being rented for, however it does say that the amount of hours is represented by T. So, fill in that information.

2. Fill in the second number.

75 + 20T = ?

Since there isn't a number to finish the second number to the equation, we know that there isn't going to be a completed answer. So, fill in the final cost with "c" as explained in the equation.

3. Write the answer.

75 + 20T = c

So, the equation to this problem is:

75 + 20T = c

Which is how to find out how much the limo costs to rent it.

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Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight

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The 95% confidence interval for the population standard deviation will be,

$$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\$

simplifying the equation, we get

$$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288$

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the 95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$ .

The 95% confidence interval for the population standard deviation will be,

$$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\$

substitute the values in the above equation, we get

$$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\$

$$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\$

simplifying the above equation, we get

$$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288$

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

brainly.com/question/14825274

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