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nordsb [41]
3 years ago
7

luis has a cooler full of soda. For every 5 cans of coke there are 3 cans of sprite. if there are 25 cans of coke, how many cans

of sprite are there?
Mathematics
1 answer:
belka [17]3 years ago
8 0

Answer: 15 cans of sprite

Step-by-step explanation:

5 cans of coke and 3 cans of sprite

then its

25 cans of coke and 15 cans of sprite

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Answer:

a) n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724  

And rounded up we have that n=6724

Step-by-step explanation:

We need to remember that the confidence interval for the true proportion is given by :  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Part a

The estimated proportion for this case is \hat p =0.76

Our interval is at 90% of confidence, and the significance level is given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. The critical values for this case are:

z_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64

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 margin of error desired is given ME =\pm 0.01 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)  

Replacing we got:

n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

Part b

For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724  

And rounded up we have that n=6724

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