If those two are your only choices then the answer is
none of the above.
<em>Attached is a cumulative frequency table for your data.</em> If you take a look at the two tables given, the frequencies were not tallied properly. If the frequency column is wrong, then the cumulative frequency will be wrong.
The answer is then none of the above or find one that matches the table attached.
Answer:
B..im pretty sure
Step-by-step explanation:
First it cant be D because 12 x __ = 64. There is no whole number multiplied by 12 that equals 64. So cross that out. then it's not B because 6 x ___ = 64. So also cross that out. the only thing u have left is 4 and 8. Then u can cross out A because the group of students has to be the same number of rows. Which is 8 x 8 aka B. The answer is B
<span>The end behavior of a polynomial function is the behavior of the graph of f as
x → +∞ or x → -∞ , depending on its leading coefficient.
In the example, the LEADING COEFFICIENT = - 3x</span>⁵
<span>,when x → +∞, ( - 3x⁵ ) → - ∞ . The end behavior, the polynomial approches minus infinity
</span>
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:
Eric's statement is false.
Step-by-step explanation:
When a positive and a negative number is added, pay attention to what number has a greater absolute value. If the positive number is greater, then the answer will be positive. In the negative number is greater, then the answer will be negative. For example, 23 + (-4) is going to end up as a positive sum, since 23 has a greater absolute value than -4. On the other hand, (-23) + 4 is going to end up as a negative number since -23 has a greater absolute value than 4.
