<h3>Equation : x + y = 170 and y = 2x - 40</h3><h3>The weight of Bill is 70 pounds and weight of mark is 100 pounds</h3>
<em><u>Solution:</u></em>
Let the weight of Bill be "x"
Let the weight of mark be "y"
Given that,
Mark and Bill have a combined weight of 170 pounds
Therefore,
x + y = 170 ------- eqn 1
Mark weighs 40 pounds less than twice Bill's weight
y = 2x - 40 ------- eqn 2
<em><u>Substitute eqn 2 in eqn 1</u></em>
x + 2x - 40 = 170
3x = 170 + 40
3x = 210
x = 70
<em><u>Substitute x = 70 in eqn 2</u></em>
y = 2(70) - 40
y = 140 - 40
y = 100
Thus weight of Bill is 70 pounds and weight of mark is 100 pounds
Answer:
9/20 or .45
Step-by-step explanation:
3/4 x 3/5 = .45 or 9/20
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
X2+y2–x–2y–11/4=x2−x+14−14+y2−2y+4−4−11/4=(x−12)2+(y−2)2−14−4−114=(x−12)2+(y−2)2−7=0(x−12)2+(y−2)2=(7√)2
Center(1/2 ,2)
radius = \sqrt 7
