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2 years ago

Which equation represents the line that passes through the points (6, –3) and (–4, –9)?

2 answers:

8 0

(6,-3)(-4,-9)
slope(m) = (-9 -(- 3) / (-4 - 6) = (-9 + 3) / -10 = -6/-10 = 3/5

y - y1 = m(x - x1)
slope(m) = 3/5
(6,-3)....x1 = 6 and y1 = -3
now we sub
y - (-3) = 3/5(x - 6) =
y + 3 = 3/5(x - 6) <==

Vesnalui [34]2 years ago

6 0

The answer is D. And for some reason this has to be at least 20 Characters so randomness.

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Twelve people each have a standard deck of cards. Each person shuffles the deck and draws two cards (without replacement). Anyon

Arada [10]

Answer:

The number of people, of the twelve, who will win a prize is 5.

Step-by-step explanation:

In a standard deck of cards there are 52 cards.

The 52 cards are divided into 4 suits: Hearts, Diamonds, Spades and Clubs, each of 13 cards.

Each of the 12 people draws 2 cards from their standard deck of cards.

Compute the probability of selecting at least 1 diamond as follows:

P (At least 1 diamond) = 1 - P (No diamond)

$=1-\frac{{39\choose 2}}{{52\choose 2}} \\=1-\frac{741}{1326}\\ =1-0.5588\\=0.4412$

The probability of selecting at least 1 diamond is 0.4412.

If a person draws at least 1 diamond he wins a prize.

The expected number of people who will win the prize is:

E (Wins a prize) = n × P (At least 1 diamond)

$=12\times0.4412\\=5.2944\\\approx5$

Thus, the number of people of the twelve who will win a prize is 5.

3 0

2 years ago

Melissa worked to earn $20.00. Alan worked for $12.00 hours. If she earns $2.00 per hour, how many hours did Melissa work?

MissTica

Answer: 10 hours

Step-by-step explanation: so if you do 10 times 2.00 it equals 20 dollars or you can divide 20 and 2

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2 years ago

If events X and Y are independent, what must be true? Check all that apply. P(Y | X) = 0 P(X | Y) = 0 P(Y | X) = P(Y) P(Y | X) =

lana [24]

if events X and Y are independent, then for intersection we multiply the probability

P(Y∩X) = P(Y) * P(X)

We know that

$P(Y|X) =\frac{P(YintersectionX)}{P(X)}$

Now we replace P(Y) * P(X) for P(Y∩X)

$P(Y|X) =\frac{P(Y)*P(X)}{P(X)}$

Cancel out P(X)

So $P(Y|X) = P(Y)$

Like that

$P(X|Y) =\frac{P(XintersectionY)}{P(Y)}$

Now we replace P(X) * P(Y) for P(X∩Y)

$P(X|Y) =\frac{P(X)*P(Y)}{P(Y)}$

Cancel out P(Y)

So $P(X|Y) = P(X)$

P(Y | X) = P(Y) and P(X | Y) = P(X) are true

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2 years ago

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It's out of B, D. I did this last year, but it's definitely out of those two choices.

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