Help please ....Stuck here

Mathematics

1 answer:

Yuki888 [10]3 years ago

4 0

Answer is C, it doesn’t pass the vertical line test. Therefore not a function

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(HMMT Combo 2009) Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expe

myrzilka [38]

The expected number of cards from the card deck that will be strictly between the two jokers is $\frac{52}{3}$ .

The most popular deck of cards used today is a normal 52-card deck of French-suited playing cards. [a] It is the only standard pack[b] used for playing cards in English-speaking nations; but, in many other nations throughout the world, it is used alongside other traditional, frequently older, standard packs with various suit systems, like those with German-, Italian-, Spanish-, or Swiss suits. The English pattern pack is the only one widely accessible in Britain and America and the most popular pattern of French-suited cards globally. The Belgian-Genoese pattern, which was created in France but is also used widely in Spain, Italy, the Ottoman Empire, the Balkans, and much of North Africa and the Middle East, is the second most popular.

Let all pairs of the positions (m,n) of the jokers are equally likely

$P[0]=P[|m-n|=1]=\frac{53}{\left(\begin{array}{c}54 \\2\end{array}\right)}=\frac{2}{27}$

$P[1]=P[|m-n|=2]=\frac{52}{\left(\begin{array}{c}54 \\2\end{array}\right)}=\frac{104}{1431}$

$P[k]=P|m-n|=k+1=\frac{54-k-1}{\left(\begin{array}{c}54 \\2\end{array}\right)}$

the expected value will be

$\begin{array}{l}\sum_{k=1}^{52} \frac{54-k-1}{\left(\begin{array}{c}54 \\2\end{array}\right)} k \\=\frac{1}{1431}\left(\sum_{k=1}^{52} 53 k-k^{2}\right) \\=\frac{1}{1431}\left(\frac{53}{2}(52)(53)-\frac{1}{6}(52)(53)(105)\right)\\=\frac{52}{3}\end{array}$