Answer:
C. ![\displaystyle 4x^2 - 20x + 26](https://tex.z-dn.net/?f=%5Cdisplaystyle%204x%5E2%20-%2020x%20%2B%2026)
Step-by-step explanation:
![\displaystyle [2x - 5]^2 + 1 = [2x - 5][2x - 5] + 1 = 4x^2 - 20x + 25 + 1 = 4x^2 - 20x + 26](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B2x%20-%205%5D%5E2%20%2B%201%20%3D%20%5B2x%20-%205%5D%5B2x%20-%205%5D%20%2B%201%20%3D%204x%5E2%20-%2020x%20%2B%2025%20%2B%201%20%3D%204x%5E2%20-%2020x%20%2B%2026)
In compositional functions, you always take the function inside parentheses FIRST, inputting that into the second function for every <em>x</em><em> </em>visible.
I am joyous to assist you anytime.
Answer:
a equals T minus b minus c minus d all over 4
Step-by-step explanation:
T is the total number of points they should have
they had 3 local matches = b c d
District match points a = 4 × (b )
T = b + c + d + a
(T - b - c - d)/4 = a
F(x) = 2ln(x) + c [integral of 1/x is ln x]
f(e^(1/2)) = 2ln(e^(1/2)) + c = 5
2(1/2) + c = 5 [ln(e^a) = a]
1 + c = 5
c = 4
So:
f(e) = 2ln(e) + 4 = 2 + 4 = 6
<span>Answer is 6.
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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:
√25 is a rational number
√1 is rational number
√7 is irrational number
Step-by-step explanation:
first one is √25
=√25
=√5² ( 25 is the square of 5)
= 5
another one is √1
= √1
=√1² ( 1 is the square of itself)
= 1
last one is √7..... and it is irrational