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3 years ago

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A football coach wants to see how many laps his players can run in 15 minutes. During a non-mandatory meeting, the coach asks fo

r volunteers on his team to do the experiment.
Which sentences explain how randomization is not applied in this situation?

Select EACH correct answer.

1 answer:

mestny [16]3 years ago

6 0

Answer:

Step-by-step explanation:

There are three reasons here why randomization is not being applied.

First, the coach is only observing football players attending the meeting. This is not representative of the entire team.

Second, there is no selection process. The coach isn't using a random generator or using a similar method.

Third, the players are volunteering for the test. The results will be biased because of this.

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An automobile club reported that the average price of regular gasoline in a certain state was $2.61 per gallon. The following d

laila [671]

Answer:

$2.618-2.98\frac{0.078}{\sqrt{15}}=2.56$

$2.618+2.98\frac{0.078}{\sqrt{15}}=2.68$

So on this case the 99% confidence interval would be given by (2.56;2.68)

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

Solution to the problem

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 mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=2.618$

The sample deviation calculated $s=0.078$

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

$df=n-1=15-1=14$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.98$

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

$2.618-2.98\frac{0.078}{\sqrt{15}}=2.56$

$2.618+2.98\frac{0.078}{\sqrt{15}}=2.68$

So on this case the 99% confidence interval would be given by (2.56;2.68)

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49 increased by d is 49 + d but I feel that this isn't the whole statement...?

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45% of c is 90. <br> Find the value of c.

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Answer:

200 = c

Step-by-step explanation:

100/45=2.2.

2.2x90= 200.

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Which explains whether or not the function represents a direct variation?

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Answer:

The function represents a direct variation

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form $y/x=k$ or $y=kx$

In a linear direct variation the line passes through the origin and the constant of proportionality k is equal to the slope m

Let

$A(0,0)$ ------> the line passes through the origin

$B(2,10)$

Find the value of k------> substitute the value of x and y

$y/x=k$ -----> $k=10/2=5$

$C(4,20)$

Find the value of k------> substitute the value of x and y

$y/x=k$ -----> $k=20/4=5$

$D(6,30)$

Find the value of k------> substitute the value of x and y

$y/x=k$ -----> $k=30/6=5$

$E(8,40)$

Find the value of k------> substitute the value of x and y

$y/x=k$ -----> $k=40/8=5$

The value of k is equal in all the points of the table and the line passes through the origin

therefore

The function represents a direct variation

the equation of the direct variation is equal to

$f(x)=5x$

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