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3 years ago

Simplify the expression 3x(x – 12x) + 3x2 – 2(x – 2)2. Which statements are true about the process and simplified product?

Mathematics

1 answer:

lora16 [44]3 years ago

6 0

Answer:

solve for x

Step-by-step explanation:

use distributed propertie and solve for x

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The n candidates for a job have been ranked 1, 2, 3,…, n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p

Anon25 [30]

Answer:

A. E(x) = 1/n×n(n+1)/2

B. E(x²) = 1/n

Step-by-step explanation:

The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as

P(x) = {1/n, x = 1,2...n}

Therefore,

Expectation of X

E(x) = summation {xP(×)}

= summation {X×1/n}

= 1/n summation{x}

= 1/n×n(n+1)/2

= n+1/2

Thus, E(x) = 1/n×n(n+1)/2

Value of E(x²)

E(x²) = summation {x²P(×)}

= summation{x²×1/n}

= 1/n

3 0

3 years ago

Find sin theta if cot theta = -2 and cos theta < 0

Katarina [22]

cot(θ) = cos(θ)/sin(θ)

So if both cot(θ) and cos(θ) are negative, that means sin(θ) must be positive.

Recall that

cot²(θ) + 1 = csc²(θ) = 1/sin²(θ)

so that

sin²(θ) = 1/(cot²(θ) + 1)

sin(θ) = 1 / √(cot²(θ) + 1)

Plug in cot(θ) = -2 and solve for sin(θ) :

sin(θ) = 1 / √((-2)² + 1)

sin(θ) = 1/√(5)

6 0

3 years ago

For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0

2 years ago

A bacteria population doubles in size every five hours after five hours the sample contains 2000 bacteria which equal the best m

Maurinko [17]

y=1000(x^2) is the answer :)

8 0

3 years ago

Theoretically, if a month is chosen 300 times,

horsena [70]

Answer:

75 times.

Step-by-step explanation:

Well there are 12 months and 3 of them start with the letter J, so theoretically, 25% of the months chosen will start with the letter J because $\frac{3}{12}=\frac{1}{4}$ , which is 25%. 25% of 300 (or $\frac{300}{4}$ , for those who like fractions) is 75, so theoretically, we can expect a month that starts with the letter J 75 times.

Hope this helps!

P.S: Please mark me as brainliest!

6 0

3 years ago