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yawa3891 [41]
4 years ago
10

Typically, joseph buys a single tire at a time for his truck. however, he sees an advertisement for a buy 3, get 1 free offer fr

om a local tire shop and decides to take advantage of the deal. joseph's decision is mostly based on
Mathematics
2 answers:
Elanso [62]4 years ago
8 0
Here is the answer that would best complete the given statement above. Typically, Joseph buys a single tire at a time for his truck. however, he sees an advertisement for a buy 3, get 1 free offer from a local tire shop and decides to take advantage of the deal. Joseph's decision is mostly based on <span>The sale price of the tires. Hope this answers your question. </span>
MA_775_DIABLO [31]4 years ago
6 0

Answer: Let's suppose that each tire costs X dollars.

then, if he buy 3, he would get 4 tires. it means that the value he gets is 4*X (because each tire cost X), and he is paying only 3*X (because he only pays for 3), which means that over time this is a better financial decision.

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Read 2 more answers
Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised t
zimovet [89]

Answer:

(b) 1/792

Step-by-step explanation:

The complete question is;

<em>Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.</em>

<em>(a) How many different groups of 5 problems can be chosen from the 12</em>

<em>problems?</em>

<em>(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?</em>

<u>In (a)</u>

<u></u>

Apply the formula

\frac{n!}{(n-r)!(r!)}

where n=12 and r=5

substitute values

=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792

In (b)

If Jerry did only 5 problems of one assignment then the probability  will be

\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}

<em />

<em />

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