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3 years ago

Find the arc measures and angles listed below.

Mathematics

1 answer:

rusak2 [61]3 years ago

4 0

Answer if you divide them or x them you'll get the answer!

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2 years ago

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3 years ago

you are renting a moving truck for a day. There is a dailyfee of $20 and a charge of $0.75 per mile. Your budget allows a maximu

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Answer:

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3 years ago

You have been assigned to determine whether more people prefer Coke or Pepsi. Assume that roughly half the population prefers Co

Alex Ar [27]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

For this case we have the following info given:

$ME=0.03$ the margin of error desired

$Conf= 0.95$ the level of confidence given

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

the critical value for 95% of confidence is $z=1.96$

We can use as estimator for the population of interest $\hat p=0.5$ . And on this case we have that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

3 0

4 years ago

Management at a seaside resort is publishing a brochure and wants to include a statement about the proportion of clear days duri

Ymorist [56]

Answer: B. 264

Step-by-step explanation:

Formula to calculate the sample size 'n' , if the prior estimate of the population proportion (p) is available:

$n= p(1-p)(\dfrac{z}{E})^2$

, where z = Critical z-value corresponds to the given confidence interval

E= margin of error

Let p be the population proportion of clear days.

As per given , we have

Prior sample size : n= 150

Number of clear days in that sample = 117

Prior estimate of the population proportion of clear days = $p=\dfrac{117}{150}$

E= 0.05

The critical z-value corresponding to 95% confidence interval = z*= 1.95 (By z-table)

Then, the required sample size will be :

$n= \dfrac{117}{150}(1-\dfrac{117}{150})(\dfrac{1.96}{0.05})^2$

Simplify ,

$n= (0.1716)(39.2)^2$

$n= 263.687424\approx264$

Hence, the sample size necessary to construct this interval =264

Thus the correct option is B. 264

7 0

3 years ago