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3 years ago

Graph each equation using its given form.

Mathematics

1 answer:

Dafna11 [192]3 years ago

8 0

For a) you would start at positive 4 then go up 2 then right 2 and repeat that

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A bag contains 5 blue balls, 4 red balls, and 3 orange balls. If a ball is picked from the bag at random, what is the probabilit

Marysya12 [62]

<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.

$BrainiacUser1357$

5 0

2 years ago

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Write (4n^4)^2 without exponents

Verdich [7]

Answer:

(4nx4nx4nx4n)x(4nx4nx4nx4n)

5 0

2 years ago

Factorise each of the following algebraic expressions completely,

LekaFEV [45]

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)

8 0

3 years ago

John and Martha are contemplating having children, but John’s brother has galactosemia (an autosomal recessive disease) and Mart

Rina8888 [55]

<h2>Answer:</h2>

$Probability=\frac{1}{24}$

<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>

$Probability=\frac{2}{3}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}\\Probability=\frac{1}{24}$

(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 $\frac{1}{24}$ </u></em>

3 0

3 years ago

Write 6,392,094,003 in word form

sattari [20]

Six billion three hundred ninety-two million ninety-four thousand and three

6 0

3 years ago

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