The answer to that would be many
In both months they had 23,703 and in January they had 13,849
Answer:
It can be concluded that the intersection of a chord and the radius that bisects it is at right angle. The two are perpendicular.
Step-by-step explanation:
i. Construct the required circle of any radius as given in the question, then locate the chord. A chord joins two points on the circumference of a circle, but not passing through its center.
ii. Construct the radius to bisect the chord, dividing it into two equal parts.
Then it would be observed that the intersection of a chord and the radius that bisects it is at right angle. Thus, the chord and radius are are perpendicular to each other.
The construction to the question is herewith attached to this answer for more clarifications.
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)
Answer & Explanation:
For this question, you need to figure out the length and the width because the area is directly related to them (A = l*w).
What we do know is that the perimeter is 36 inches. Since the perimeter is equal to P = 2l + 2w, we can say:
P = 36 = 2(3 + 2w) + 2(w), where the length l is 3 + 2w
Solve for w first, then solve for l. Use these values to find the area.
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