<u>Answer:</u>
- There is a 41.7% chance that you got a blue ball.
<u>Step-by-step explanation:</u>
<u>Total number of blue balls/Total number of balls x 100 = Chance of getting a blue ball</u>
- => 5/12 x 100
- => 5/6 x 50
- => 5/3 x 25
- => 125/3
- => 41.7
Hence, <u>there is a 41.7% chance that you got a blue ball.</u>
Hoped this helped.

Answer:
(4nx4nx4nx4n)x(4nx4nx4nx4n)
Answer:
see explanation
Step-by-step explanation:
(a)
Given
2k - 6k² + 4k³ ← factor out 2k from each term
= 2k(1 - 3k + 2k²)
To factor the quadratic
Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)
The factors are - 1 and - 2
Use these factors to split the k- term
1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )
1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term
= (1 - k)(1 - 2k)
1 - 3k + 2k² = (1 - k)(1 - 2k) and
2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)
(b)
Given
2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )
= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term
= (x - 2y)(2a + 3b)
<h2>
Answer:</h2>

<h2>
Step-by-step explanation:</h2>
As the question states,
John's brother has Galactosemia which states that his parents were both the carriers.
Therefore, the chances for the John to have the disease is = 2/3
Now,
Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.
Now, one of those children married with a person.
So,
Probability for the child to have disease will be = 1/2
Now, again the child's child (Martha) probability for having the disease is = 1/2.
Therefore,
<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)
<em><u>Hence, the probability for the first child to have Galactosemia is
</u></em>
Six billion three hundred ninety-two million ninety-four thousand and three