Let n represent the number. You require
7n - 2n = 55
5n = 55 . . . . . . collect terms
n = 11 . . . . . . . divide by 5
The number is 11.
Step-by-step explanation:
Number of males is
Number of males not enrolled
Number of females not enrolled
(a)
The table based on data is
<u> Enrolled Not enrolled Total </u>
<u>Male 82 36 118 </u>
<u>Female 102 56 158 </u>
Total 184 92 276
(b)
Percentage of students were males that went to magic college
- enrolled male / total students = (use table above)
- 82/276*100% = 29.71% (rounded)
(c)
Percentage of females went to magic college
- enrolled female / total female = (use table above)
- 102/158*100% = 64.56% (rounded)
Step-by-step explanation:
From this figure TUW+WUV=TUV ; (5x+3)+(10x-5)=17x-16 so 15x-2=17x-16 ;14=2x therefore x=7
then substitute x on it.
TUW=5x+3=5(7)+3=35+3=38
WUV=10x-5=10(7)-5=70-5=65
TUV=17x-16=17(7)-16=119-16=103
Order : 6 9 9 10 12 15 15 17 18
Mean: 12.33
Median: 12
mode: 9 and 15
range: 12
Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way
Step-by-step explanation:
- 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.
- 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)
