Answer:
firstable give c+b a polynomial value like x
so its will be x^2+11x+30
after the we have to factor it
30=6×5
and 11=6+5
so its will become
(x+6)×(x+5)=x^2+11x+30
x=c+d
(c+d+6)×(c+d+5)=(c+d)^2+11(c+d)+30
have a great day
Answer:
- B. The probability that the student received a grade better than C is 0.45
Step-by-step explanation:
<u>Total number of students:</u>
- 15 + 30 + 35 + 16 + 4 = 100
<h3>Statements</h3>
A.
- p(<D) = 4/100 = 0.04
- 0.04 ≠ 0.2
- Incorrect
B.
- p(>C) = (30 + 15)/100 = 45/100 = 0.45
- 0.45 = 0.45
- Correct
C.
- p(not A or B) = (100 - 15 - 30)/100 = 55/100 = 0.55
- 0.55 ≠ 0.45
- Incorrect
D.
- p(B or C) = (30 + 35)/100 = 65/100 = 0.65
- 0.65 ≠ 0.35
- Incorrect
<h3>
Answer:</h3>
1/17 or 0.0588 (without replacement)
<h3>
Step-by-step explanation:</h3>
To answer this question we need to know the following about a deck of cards
- A deck of cards contains 4 of each card (4 Aces, 4 Kings, 4 Queens, etc.)
- Also there are 4 suits (Clubs, Hearts, Diamonds, and Spades).
- Additionally, there are 13 cards in each suit (Clubs/Spades are black, Hearts/Diamonds are red)
.
In this case, we are required to determine the probability of choosing two diamonds.
- There are 13 diamonds in the deck.
- Assuming, the cards were chosen without replacement;
P(Both cards are diamonds) = P(first card is diamond) × P(second card is diamond)
P(First card is diamond) = 13/52
If there was no replacement, then after picking the first diamond card, there are 12 diamond cards remaining and a total of 51 cards remaining in the deck.
Therefore;
P(Second card is diamond) = 12/51
Thus;
P(Both cards are diamonds) = 13/52 × 12/51
= 156/2652
= 1/17 or 0.0588
Hence, the probability of choosing two diamonds at random (without replacement) is 1/17 or 0.0588.
The solution is -10 because it’s absolute value
Answer:
5,680.44
Step-by-step explanation:
