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Aliun [14]
3 years ago
13

I need help, what do I do?

Mathematics
1 answer:
AnnZ [28]3 years ago
6 0
I'm sorry sorry don't no understand brother
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A test consists of 490 true or false questions. If the student guesses on each question, what is the standard deviation of the n
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I believe the answer is 1.
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Is this correct btw I put 1400​
Oxana [17]

Answer: Yes

Step-by-step explanation: The answer of 1400 is correct.

What I did is I found the area of the figure assuming it was a rectangle, then I subtracted the corners that were removed.

1800 - 150 - 100 - 150 = 1400

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Y=-2/3x+2 <br> Y=3x-4<br> Y=4x-6<br> Algebra 1
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3 years ago
Maria and Jason are saving gas money for their trip to Boston. Maria has saved $20 more than double what Jason has. Combined Mar
valentina_108 [34]

Answer:

$60

Step-by-step explanation:

Jason saved gas money (x).  Maria saved $20 more than double what Jason has (20 + 2x).  Combined, they saved $200.  Solve for x for Jason's amount.

(x) + (20 + 2x) = 200

x + 20 + 2x = 200

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3x + 20 = 200

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Jason saved $60 for this trip.

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3 years ago
Crash Hospital Costs. A study was conducts to estimate hospital costs for accident victims who wore seat belts. Twenty randomly
kvv77 [185]

Answer:

a) The 99% confidence interval would be given by (5004.168;12603.832)

b) For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I

Step-by-step explanation:

1) 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 =9004 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=5629 represent the sample standard deviation

n=20 represent the sample size  

2) 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)

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

df=n-1=20-1=19

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(0.005,19)".And we see that t_{\alpha/2}=2.86

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

9004-2.86\frac{5629}{\sqrt{20}}=5004.168    

9004+2.86\frac{5629}{\sqrt{20}}=12603.832

So on this case the 99% confidence interval would be given by (5004.168;12603.832)    

3) Part b

For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I

7 0
3 years ago
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