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lapo4ka [179]
3 years ago
9

Correct/best answer will get brainliest!

Mathematics
1 answer:
vodka [1.7K]3 years ago
6 0
The answer would be b.100 because its clearly not a right triangle
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Assume that Emily (who has CF, a recessive disease (aa)) decides to have children with a man who does not have CF and who has no
alukav5142 [94]

Answer:100%. Probability

Step-by-step explanation:

Emily = aa

Man = AA

After crossing the two, the result wI'll be Aa, Aa, Aa, Aa

Since every result has a carrier(a) in it, makes it 100% probability.

7 0
3 years ago
Select the equation and solution for the problem. Let x represent the unknown number.
Alinara [238K]
Answer

I am positive it us 4x = 14 I truly apologize if it's not.
Step-by-step explanation

8 0
2 years ago
Larry had 3⁄7 of a strawberry-rhubarb pie left over. He split the leftover pie evenly between his 3 children. What fraction of a
Damm [24]

Answer:

1/7

Step-by-step explanation:

The left over fraction is divided among the three children as follows

Given

Left over as 3/7

Number of children 3

Quantity each gets will be 3/7÷3

3/7÷3=3/7*1/3=3/21

By simplyfying the fractions, 3/21 when both the numerator and denominator are divided by 3 we obtain that 3/21=1/7

Therefore, each child geta 1/7

7 0
3 years ago
Ty in advance for whoever does this. Find the range and put it on the graph. thank you <3
ser-zykov [4K]

Answer:

Step-by-step explanation:

y_{1} = - 4 - 2( - 3) = 2

y_{2} = - 4 - 2( - 2) = 0

y_{3} = - 4 - 2( - 1) = - 2

y_{4} = - 4 - 2( 0 ) = - 4

y_{5} = - 4 - 2( 1 ) = - 6

Range: { 2, 0, - 2, - 4, - 6 }

8 0
3 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
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