What is the numerator of the simplified sum? StartFraction x Over x squared + 3 x + 2 EndFraction + StartFraction 3 Over x + 1 E
ndFraction
1 answer:
Answer:
4x + 6
Step-by-step explanation:
Given
+
Before we can add the fractions we require them to have a common denominator.
Factor the denominator of the first fraction
+
Multiply the numerator / denominator of the second fraction by (x + 2)
= +
← fractions now have a common denominator
Add the numerators leaving the denominators
=
=
= ← simplified sum with numerator 4x + 6
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Answer:
Step-by-step explanation:
<u>Theorem 1:</u> The height of right triangle drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of these two segments. Hence,
<u>Theorem 2:</u> In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.
Thus,
Answer:
The probabilities of each outcome are the following:
for X = 1 is 0.6
for X = 2 is 0.35
for X = 3 is 0.049
and for X = 4 is 0.001
Step-by-step explanation:
Let's consider X as the random variable for the sum of outcomes "S" exceeds 3, this is:
Let's now analyze and consider the ways that X equals the different values:
X = 1: I throw the dice and the result is: 4, 5 and 6.
Then: P(X=1) = P(4) + P(5) + P(6) = 0.2+0.1+0.3 = 0.6
Thus P(X=1) = 0.6
X = 2: The result after following the dice two times can be:
1 and 3, 1 an 4.... and so on until 1 and 6
2 and 2, 2 and 3... and so on until 2 and 6
3 and 1... and so on until 3 a 6
Then P(X=2) = P(1)xP(3,4,5,6) + P(2)xP(2....6) + P(3)x(1.....6)
Theory of Probability: Sum of all possible outcomes P(1)+P(2).......P(6) = 1
Then P(1......6) = 1
Then P(X=2) = P(1)x[1-P(1)-P(2)]+P(2)x[1-P(1)]+P(3) = 0.1x(1-0.1-0.2) + 0.2x(1-0.1) + 0.1 = 0.1 x 0.7 + 0.2 x 0.9 + 0.1 = 0.35
Thus P(X=2) = 0.35
X = 3: The result can be
1 and 1 and 2, 1 and 1 and 3.... until 1 and 1 and 6
1 and 2 and 1, 1 and 2 and 2..... until 1 and 2 and 6
2 and 1 and 1, 2 and 1 and 2.... until 2 and 1 and 6
Then P(X=3) = P(1)xP(1)xP(2....6) + P(1)xP(2)xP(1.....6) + P(1)xP(2)xP(1.....6)
P(X=3) = 0.1 x 0.1 x (1-0.1) + 0.1 x 0.2 x 1 + 0.2 x 0.1 x 1 = 0.01 x 0.9 + 0.2 + 0.2 = 0.049
Thus P(X=3) = 0.049
Finally, for X to be 4, I only have the following possibilities
1 and 1 and 1 and 1.... until 1 and 1 and 1 and 6
Then P(X=4) = P(1)xP(1)xP(1)xP(1.....6) = 0.1x0.1x0.1x1 = 0.001
Thus P(X=3) = 0.001