All categories

2 years ago

Which points are coplanar? A. C,G,A,F B. A,J,K,I C. B,J,K,F D. A,B,H,G

Mathematics

1 answer:

Damm [24]2 years ago

4 0

Answer:

B, J, K, F. Because they are on the same side compared to the other answers.

Step-by-step explanation:

Coplanar points are three or more points which all lie in the same plane. Any set of three points in space is coplanar. A set of four points may be coplanar or may be not coplanar.

Plz mark brainliest!

You might be interested in

In a dog show, there will be a gold and silver medal awarded.There are 6 finalists. In how many ways can the 2 medals be awarded

svp [43]

Answer:

(6)(5) =

Step-by-step explanation:

3 0

3 years ago

Please explain as well 3x+5y=21 4x-2y=-244x−2y=−24

hodyreva [135]

Step-by-step explanation:

3x + 5y = 21 * (-4) •••••• -12x -20y = -84 (a)

4x - 2y = -24 * (3) •••••• 12x -6y = -72 (b)

(a) + (b)

-26y = -156

y = 6

3x + 5 * 6 = 21

3x = 21 - 30

3x = -9

x = -3

3 0

3 years ago

I NEED HELP FAST!! PLZ.

adoni [48]

The answer is z = 4 because of you put 4 into the equation it equals to 100

3 0

2 years ago

Is this right please help me Is this right please help me

saw5 [17]

Yes you are correct

7 0

2 years ago

The probability that a rental car will be stolen is. 4. if 3500 cars are rented, what is the approximate poisson probability tha

kozerog [31]

Using the Poisson distribution, there is a 0.8335 = 83.35% probability that 2 or fewer will be stolen.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The probability that a rental car will be stolen is 0.0004, hence, for 3500 cars, the mean is:

$\mu = 3500 \times 0.0004 = 1.4$

The probability that 2 or fewer cars will be stolen is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.4}1.4^{0}}{(0)!} = 0.2466$

$P(X = 1) = \frac{e^{-1.4}1.4^{1}}{(1)!} = 0.3452$

$P(X = 2) = \frac{e^{-1.4}1.4^{2}}{(2)!} = 0.2417$

Then:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2466 + 0.3452 + 0.2417 = 0.8335$

0.8335 = 83.35% probability that 2 or fewer will be stolen.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

4 0

1 year ago