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3 years ago

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Which of the following are point-slope equations of the line going through (-3, 4) and (2, 1)? Check all that apply. A. y-1 = g(

x-2) B. y-4 = (x+3) O C. y-4 = -(x+3) D. Y+ 2 = (x-1 E. Y+3 = (x-4) OF. y-1 = -(x-2)

1 answer:

Artist 52 [7]3 years ago

5 0

The point slope equation of the line going through (-3,4) and (2,1) is

$y - 1 = - 3 \frac{(x - 2)}{5}$

or,

$y - 4 = - 3 \frac{(x + 3)}{5}$

Answered by GAUTHMATH

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<h2>11.f(x)=^2-3</h2>

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<h2>12.And slide everything five units down. So a over here that's negative three negative one. So if we slide at five you just down we go one two three four five and get here.</h2>

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3 years ago

Simplify -2.9f + .9f -12 - 4

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3 years ago

(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

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

<h3>What is card deck?</h3>

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

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

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the expected value will be

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

Therefore, the expected number of cards that will be strictly between the two jokers is $\frac{52}{3}$ .

To learn more about card deck from the given link:

brainly.com/question/13477028

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2 years ago

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