We can use the fact that, for 
,
Notice that
![\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{1-x}\right]=\dfrac1{(1-x)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1%7B1-x%7D%5Cright%5D%3D%5Cdfrac1%7B%281-x%29%5E2%7D)
so that
![f(x)=\displaystyle\frac5{(1-x)^2}=5\frac{\mathrm d}{\mathrm dx}\left[\sum_{n=0}^\infty x^n\right]](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdisplaystyle%5Cfrac5%7B%281-x%29%5E2%7D%3D5%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20x%5En%5Cright%5D)
By the ratio test, this series converges if
so the series has radius of convergence 
.
Answer:
D
Step-by-step explanation:
The required value of y is 2645 / 77 when x = 77.
Step-by-step explanation:
1. Let's check the information given to answer:
if y varies inversely as x, and y = 23 when x = 115. Find y when x=77
2. What is y when x = 77?
As the statement "y varies inversely as x" translates into y = k/x
If we let y = 23 and x = 115, the constant of variation becomes
23 = k ÷ 115
k = 23 × 115
k = 2645
Thus the specification equation is y = 2645 ÷ x . Now, letting x = 77, we obtain
y = 2645 ÷ x
y = 2645 ÷ 77
<u>So, The required value of y = 2645 / 77 when x = 77</u>
<u></u>
<u>#learnwithBrainly</u>
From a standard deck of cards, one card is drawn. What is the probability that the card is black and a
jack? P(Black and Jack)
P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26
A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen
or an ace. P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13
WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the
probability that they will both be aces? P(AA) = (4/52)(3/51) = 1/221.
1
WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a
king? P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been
removed.
WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick
a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the
probability of drawing the first queen which is 4/52. The probability of drawing the second queen is also
4/52 and the third is 4/52. We multiply these three individual probabilities together to get P(QQQ) =
P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.
Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)
What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note:
A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with
13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant,
and you keep each card as it is dealt -- it's not returned to the deck.)
The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken
from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and
there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the
first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for
the other three. The probability of the royal flush is therefore the product of these numbers, or
5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154
Here you go I Found a graph for you
