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Alisiya [41]
3 years ago
10

Joanie runs the concession stand for the school's baseball games. At each game, the best selling items are pretzels and hot dogs

. Pretzels are $3 each, and hot dogs are $2 each. Today she sold 15 more hot dogs than pretzels and made $195 in total sales.
Use a system of equations to model the situation above, and determine which of the following are possible amounts of pretzels and hot dogs that Joanie sold today.

A. 35 pretzels, 45 hot dogs

B. 15 pretzels, 30 hot dogs

C. 31 pretzels, 51 hot dogs

D. 33 pretzels, 48 hot dogs
Mathematics
1 answer:
Ronch [10]3 years ago
6 0

The answer is D. 33 pretzels, 48 hot dogs

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My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the
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Answer:

a) n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07

And rounded up we have that n=1476

b) n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

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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)  

If solve n from equation (a) we got:  

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

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In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.9=0.1 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=\pm 1.64  

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

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And on this case we have that ME =\pm 0.02 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{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07

And rounded up we have that n=1476

Part b

For this case since we don't have a prior estimate we can use \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

And rounded up we have that n=1681

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