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3 years ago

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Each student on the trip to cooperstown paid the same amount .the tottal amount collected was $589.pilar paid for her ticket usi

ng exactly three bils . how much did each ticket coast?
Please help

1 answer:

professor190 [17]3 years ago

5 0

Answer:

$231.67

Step-by-step explanation:

how you need to divide 589 to 3 and than you get 231.67

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Answer:

2/5

Step-by-step explanation:

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Stephanie has a total of 27 nickels and dimes. The total value of the coins is $1.95. Which system of equations can be used to f

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3 years ago

Select the correct answer from the drop-down menu.

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Answer:

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Step-by-step explanation:

Use FOIL method

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3 years ago

A country clerk has given amount of money to budget for cultural events

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Answer:

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Step-by-step explanation:

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3 years ago

One step in the manufacture of a certain metal clamp involves the drilling of four holes. In a sample of 150 clamps, the average

Mandarinka [93]

Answer:

a) $72-1.52\frac{10}{\sqrt{150}}=70.76$

$72+1.52\frac{10}{\sqrt{150}}=73.24$

So on this case the 87% confidence interval would be given by (70.76;73.24)

b) $ME= \frac{75-69}{2}= 3$

And the margin of error is given by:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we can solve for the critical value $t_{\alpha/2}$

And we got:

$t_{\alpha/2}= \frac{ME*\sqrt{n}}{s}= \frac{3* \sqrt{150}}{10}= 3.674$

And we can find the significance level with this formula:

"=2*T.DIST(-3.674;149;TRUE)"

And the confidence level would be 1-0.000332= 0.999668

And then that correspond to 99.96%

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".

$\bar X=72$ represent the sample mean

$\mu$ population mean (variable of interest)

$s=10$ represent the sample standard deviation

n=150 represent the sample size

87% confidence interval

Part a

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

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

The degrees of freedom are given by:

$df = 150-1=149$

Since the Confidence is 0.87 or 87%, the value of $\alpha=0.13$ and $\alpha/2 =0.065$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.065,149)".And we see that $z_{\alpha/2}=1.52$

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

$72-1.52\frac{10}{\sqrt{150}}=70.76$

$72+1.52\frac{10}{\sqrt{150}}=73.24$

So on this case the 87% confidence interval would be given by (70.76;73.24)

Part b

For this case the confidence interval is given by: (69,75)

We can estimate the margin of error like this:

$ME= \frac{75-69}{2}= 3$

And the margin of error is given by:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we can solve for the critical value $t_{\alpha/2}$

And we got:

$t_{\alpha/2}= \frac{ME*\sqrt{n}}{s}= \frac{3* \sqrt{150}}{10}= 3.674$

And we can find the significance level with this formula:

"=2*T.DIST(-3.674;149;TRUE)"

And the confidence level would be 1-0.000332= 0.999668

And then that correspond to 99.96%

3 0

3 years ago