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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?

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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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

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