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3 years ago

Multiply: (6x - 1)(6x +1)

Mathematics

2 answers:

scoundrel [369]3 years ago

6 0

Answer:

36x^2 -1

Step-by-step explanation:

6x times 6x= 36x^2

6x times 1= 6x

-1 times 6x= -6x

-1 times 1= -1

so, ...

36x^2+6x+(-6x)+(-1) = 36x^2 -1

Salsk061 [2.6K]3 years ago

4 0

I’m not positive but i think it’s 36x^2 -1

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The two sides of a triangle have lengths 4 and 10. The third side of the triangle can be?

spin [16.1K]

Answer:

10.77

Step-by-step explanation:

If you use the pythagorean theorem to find the hypotenuse, you should get 10.77

4 0

3 years ago

A computer retail store has 1414 personal computers in stock. A buyer wants to purchase 33 of them. Unknown to either the retail

Lorico [155]

Answer:

a) 364 ways

b) 45.33% probability that exactly one of the computers will be defective.

c) 54.67% probability that at least one of the computers selected is defective.

Step-by-step explanation:

The computers are chosen without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

14 computers, so N = 14.

3 defective, so k = 3.

3 will be purchesed, so n = 3.

A) In how many different ways can the 3 computers be chosen?

3 from a set of 14. So

$C_{14,3} = \frac{14!}{3!(14-3)!} = 364$

364 ways

B) What is the probability that exactly one of thecomputers will be defective?

This is P(X = 1).

$P(X = 1) = h(1,14,3,3) = \frac{C_{3,1}*C_{11,2}}{C_{14,3}} = 0.4533$

45.33% probability that exactly one of the computers will be defective.

C) What is the probability that at least one of the computers selected is defective?

Either none is, or at least one is. The sum of the probabilities of these events is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = h(0,14,3,3) = \frac{C_{3,0}*C_{11,3}}{C_{14,3}} = 0.4533$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4533 = 0.5467$

54.67% probability that at least one of the computers selected is defective.

5 0

3 years ago

Which figure has the correct lines of symmetry drawn in?

RoseWind [281]

Its actually the first one because its a hexagon and that line Is the only line that can cut the shape in half

4 0

4 years ago

Read 2 more answers

A survey asked high school cell phone users how many minutes they use their phones to talk each day. The

kicyunya [14]

The two numbers in between which the lower quartile value given by the attached boxplot falls are 1.5 and 2.5 respectively.

The lower quartile means the 25th percentile or lower $\frac{1}{4}th$ of the distribution

From a boxplot, the lower quartile is the point marked by the starting point of the box.

The Lower quartile of the distribution represented by the boxplot is 2

The values which bounds the lower quartile value to the left and right are 1.5 and 2.5 respectively.

Therefore, two numbers in between which the lower quartile value falls are 1.5 and 2.5

Learn more :brainly.com/question/24582786

6 0

3 years ago

Let be the statement ""x has a cat,"" let be the statement ""x has a dog,"" and let be the statement ""x has a ferret."" Express

jarptica [38.1K]

Complete question is:

Let C(x) be the statement “x has a cat,” let D(x) be the statement “x has a dog”, and let F(x) be the statement “x has a ferret”. Express each of these statements in terms of C(x), D(x), F(x), quantifiers, and logical connectives. Let the domain consist of all students in your class.

a) A student in your class has a cat, a dog, and a ferret.

b) All students in your class have a cat, a dog, or a ferret.

c) Some student in your class has a cat and a ferret, but not a dog.

d) No student in your class has a cat, a dog, and a ferret.

e) For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.

Answer:

a) ∃x(C(x) ∧ D(x) ∧ F(x))

b) ∀x(C(x) ∨ D(x) ∨ F(x))

c) ∃x(C(x) ∧ D(x) ¬ F(x))

d) ¬∃x(C(x) ∧ D(x) ∧ F(x))

e) (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))

Step-by-step explanation:

We are given;

C(x) = “x has a cat”

D(x) = “x has a dog”

F(x) = “x has a ferret”

We will make use of the following interpretation symbols;

Negation ¬ A : Not A

Disjunction A ∨ B : A or B

Conjunction A ∧ B : A and B

Existential quantification ∃ x A(x) : There exists an element x in the domain such that A(x)

Universal Quantification ∀x A(x) : A(x) for all values of x in the domain

So, using the above interpretations, we can answer the questions as;

a) ∃x(C(x) ∧ D(x) ∧ F(x))

b) ∀x(C(x) ∨ D(x) ∨ F(x))

c) ∃x(C(x) ∧ D(x) ¬ F(x))

d) ¬∃x(C(x) ∧ D(x) ∧ F(x))

e) (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))

8 0

3 years ago