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lys-0071 [83]
3 years ago
5

A local hamburger shop sold a combined total of 500 hamburgers and cheeseburgers on Wednesday. There were 50 fewer cheeseburgers

sold than hamburgers. How many hamburgers were sold on Wednesday?
Mathematics
1 answer:
Virty [35]3 years ago
6 0

Hey there! I'm happy to help!

Let's call the hamburgers h and the cheeseburgers c.

h+c=500

c=h-50

Let's plug this value for c into the first equation to solve for h.

h+h-50=500

2h-50=500

Add 50 to both sides.

2h=550

Divide both sides by 2.

h=275

Therefore, 275 hamburgers were sold on Wednesday.

Have a wonderful day! :D

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Given the series \sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}

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\frac{a_n_+_1}{a_n} =  \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\

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|\frac{1+2/\infty+1/\infty^2}{5}|\\\\

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d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

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Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

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then N≥(\frac{1.96*50}{8} )^2 ≈150.6

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