52 quadrilaterals would have 52 ÷ 4 = a
a = your answer.
My answer is reasonable because if you have 52 sides from quadrilaterals you would need to divide by 4 to get the amount of quadrilaterals you have. Check your work by multiplying 4 x a = __ (The blank should be 52)

5 0

3 years ago

In △RST, RW=27 cm.

sertanlavr [38]

Hello!

I just completed this test and if I'm correct in assuming that this is the same problem I did, then your answer would be 18 cm.

Hope this helps! :)

7 0

2 years ago

There are 4 green marbles and 2 red marbles in the jar. You just randomly draw one by one without replacement and stop when you

zalisa [80]

Answer:

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3) = 2/15

P(4) = 1/15

The expected value for X is:

EV = 1.33...

Step-by-step explanation:

We have a total of 6 marbles in the jar.

The probability of getting a red marble in the first try (X = 0) is equal to the quotient between the number of red marbles and the total number of marbles, this is:

P(0) = 2/6 = 1/3

The probability of drawing one green marble (X = 1)

is:

First, you draw a green marble with a probability of 4/6

Then you draw the red one, but now there are 5 marbles in the jar (2 red ones and 3 green ones), then the probability is 2/5

The joint probability is:

P(1) = (4/6)*(2/5) = (2/3)*(2/5) = 4/15

The probability of drawing two green marbles (X = 2)

Again, first we draw a green marble with a probability of 4/6

Now we draw again a green marble, now there are 3 green marbles and 5 total marbles in the jar, so this time the probability is 3/5

Now we draw the red marble (there are 2 red marbles and 4 total marbles in the jar), with a probability of 2/4

The joint probability is:

P(2) = (4/6)*(3/5)*(2/4) = (2/6)*(3/5) = 1/5

The probability of drawing 3 green marbles (X = 3)

At this point you may already understand the pattern:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

finally, we draw a red marble with a probability 2/3

The joint probability is:

P(3) = (4/6)*(3/5)*(2/4)*(2/3) = (2/6)*(3/5)*(2/3) = (1/5)*(2/3) = (2/15)

Finally, the probability of drawing four green marbles (X = 4) is given by:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

fourth, we draw a green marble with a probability 1/3

Finally, we draw a red marble with a probability 2/2 = 1

The joint probability is:

P(4) = (4/6)*(3/5)*(2/4)*(1/3)*1 = (1/5)*(1/3) = 1/15

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3) = 2/15

P(4) = 1/15

The expected value will be:

EV = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3) + 4*P(4)

EV = 1*(4/15) + 2*( 1/5) + 3*( 2/15) + 4*(1/15 ) = 1.33

So we can expect to draw 1.33 green marbles in this experiment.

5 0

3 years ago