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eimsori [14]
3 years ago
5

A footballer scores in one out of five matches. What percentage of the matches does the footballer score in?

Mathematics
1 answer:
MaRussiya [10]3 years ago
7 0

Answer:

20%

Step-by-step explanation:

100 divided by 5 =20 20x1=20

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Question 13 PLEASE HELP Find the x-intercept of -x +2y =6. X-intercept: (___,0)
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Step-by-step explanation:

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A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n
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Answer:

i) 2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959  

2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041  

So then the 90% confidence interval is given by (1.959, 2.041)

ii) Figure attached

iii) ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041

Step-by-step explanation:

Previous concepts

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".

The confidence interval is given by this formula:

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

And for a 90% of confidence the significance is given by \alpha=1-0.9=0.1, and \frac{\alpha}{2}=0.05. Since we know the population standard deviation we can calculate the critical value z_{0.05}= \pm 1.64

We know the folllowing data:

\bar X = 2 represent the sample mean

\sigma = 0.1 represent the population deviation

n =16 represent the sample size

Part i)

If we replace the values given into formula (1) we got:

2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959  

2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041  

So then the 90% confidence interval is given by (1.959, 2.041)

Part ii)

Figure attached. We have the illustration for the confidence interval obtained.

Part iii)

The margin of error is given by:

ME=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}}

And if we replace we got:

ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041

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