The factor of the expression 8r⁶ + 27s¹² can be (2r²+3s⁴)(4r⁴-6r²s⁴+9s⁸) after using the identity a³+b³ = (a+b)(a²-ab+b²).
It is defined as the combination of constants and variables with mathematical operators.
It is given that:
= 8r⁶ + 27s¹²
The above expression can be written as:
8r⁶ + 27s¹² = (2r²)³+(3s⁴)³
After applying the identity:
a³+b³ = (a+b)(a²-ab+b²)
= (2r²+3s⁴)(4r⁴-6r²s⁴+9s⁸)
Thus, the factor of the expression 8r⁶ + 27s¹² can be (2r²+3s⁴)(4r⁴-6r²s⁴+9s⁸) after using the identity a³+b³ = (a+b)(a²-ab+b²).
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Answer:
a^3+b^3=(a+b)(a²-ab+b²)
8r^6+27s^12=(2r²)^3+(3s^4)^3=(2r²+3s^4)(4r^4-6r²*s^4+9s^8)
x = –1
No Real Solutions
x = 11 and x = 1
x = 11
The solution to the quadratic equation x² - 12x + 11 = 0 is x = 11 and x = 1.
Given the quadratic equation in the question:
x² - 12x + 11 = 0
To solve the quadratic equation x² - 12x + 11 = 0 by completing the square, first the constant term to the other side of the equation:
x² - 12x + 11 = 0
x² - 12x + 11 - 11 = 0 - 11
x² - 12x = -11
Next, find the value that is equal to the square of half of b:
( b/2 )² = ( -12/2 )² = (-6)²
Add (-6)² to each side of the equation:
x² - 12x + (-6)² = -11 + (-6)²
x² - 12x + 36 = -11 + 36
x² - 12x + 36 = 25
Factor the perfect trinomial sqaure:
( x - 6 )² = 25
Solve for x:
x - 6 = ±√25
x - 6 = ±5
x = 6 ± 5
Hence, x = 6 - 5 = 1
And x = 6 + 5 = 11.
Therefore, the values of x are 1 and 11.
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Answer:
Step-by-step explanation:
x^2 – 12x + 11 = 0
(x-11)(x-1)
x= 11, 1 completing the square
Answer:
a) the probability of waiting more than 10 min is 2/3 ≈ 66,67%
b) the probability of waiting more than 10 min, knowing that you already waited 15 min is 5/15 ≈ 33,33%
Step-by-step explanation:
to calculate, we will use the uniform distribution function:
p(c≤X≤d)= (d-c)/(B-A) , for A≤x≤B
where p(c≤X≤d) is the probability that the variable is between the values c and d. B is the maximum value possible and A is the minimum value possible.
In our case the random variable X= waiting time for the bus, and therefore
B= 30 min (maximum waiting time, it arrives 10:30 a.m)
A= 0 (minimum waiting time, it arrives 10:00 a.m )
a) the probability that the waiting time is longer than 10 minutes:
c=10 min , d=B=30 min --> waiting time X between 10 and 30 minutes
p(10 min≤X≤30 min) = (30 min - 10 min) / (30 min - 0 min) = 20/30=2/3 ≈ 66,67%
a) the probability that 10 minutes or more are needed to wait starting from 10:15 , is the same that saying that the waiting time is greater than 25 min (X≥25 min) knowing that you have waited 15 min (X≥15 min). This is written as P(X≥25 | X≥15 ). To calculate it the theorem of Bayes is used
P(A | B )= P(A ∩ B ) / P(A) . where P(A | B ) is the probability that A happen , knowing that B already happened. And P(A ∩ B ) is the probability that both A and B happen.
In our case:
P(X≥25 | X≥15 )= P(X≥25 ∩ X≥15 ) / P(X≥15 ) = P(X≥25) / P(X≥15) ,
Note: P(X≥25 ∩ X≥15 )= P(X≥25) because if you wait more than 25 minutes, you are already waiting more than 15 minutes
- P(X≥25) is the probability that waiting time is greater than 25 min
c=25 min , d=B=30 min --> waiting time X between 25 and 30 minutes
p(25 min≤X≤30 min) = (30 min - 25 min) / (30 min - 0 min) = 5/30 ≈ 16,67%
- P(X≥15) is the probability that waiting time is greater than 15 min --> p(15 min≤X≤30 min) = (30 min - 15 min) / (30 min - 0 min) = 15/30
therefore
P(X≥25 | X≥15 )= P(X≥25) / P(X≥15) = (5/30) / (15/30) =5/15=1/3 ≈ 33,33%
Note:
P(X≥25 | X≥15 )≈ 33,33% ≥ P(X≥25) ≈ 16,67% since we know that the bus did not arrive the first 15 minutes and therefore is more likely that the actual waiting time could be in the 25 min - 30 min range (10:25-10:30).
Answer:
1/34
Step-by-step explanation:
To find the probability that a specific number is chosen
P ( number 13) = 1/ total number of balls
= 1 /34
Answer:
1/34
Step-by-step explanation:
There is one 13 ball and there are 34 balls total.
1/34 is the answer. :)