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spayn [35]
3 years ago
15

A spinner has 10 equally sized sections, 4 of which are gray and 6 of which are blue. The spinner is spun twice. What is the pro

bability that the first spin lands
on blue and the second spin lands on gray?
Mathematics
1 answer:
wariber [46]3 years ago
4 0

Answer:

1/2 beacuse its each have a possibility tobe land on and the most best answer wpuld be 1/2 i tryed ok

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Mazyrski [523]

Answer:

a) 654.16-2.01\frac{165.4}{\sqrt{52}}=608.06    

654.16+2.01\frac{165.4}{\sqrt{52}}=700.26    

b) n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189

So the answer for this case would be n=189 rounded up to the nearest integer

Step-by-step explanation:

Part a

\bar X=654.16 represent the sample mean

\mu population mean (variable of interest)

s=165.4 represent the sample standard deviation

n =52represent the sample size  

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 aregiven by:

df=n-1=52-1=51

Since the Confidence is 0.95 or 95%, the significance \alpha=0.05 and \alpha/2 =0.025, and the critical value would be t_{\alpha/2}=2.01

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

654.16-2.01\frac{165.4}{\sqrt{52}}=608.06    

654.16+2.01\frac{165.4}{\sqrt{52}}=700.26    

Part b

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (b)

The critical value for this case wuld be z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189

So the answer for this case would be n=189 rounded up to the nearest integer

7 0
3 years ago
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