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svetoff [14.1K]
2 years ago
13

Desperate for helllppp pls

Mathematics
1 answer:
beks73 [17]2 years ago
7 0

Answer:

  • B) - 1 < x < - 0.2

Step-by-step explanation:

<h3>Given inequality</h3>

  • |5x + 3| < 2

The LHS is absolute value so we look at two options:

<h3>Option 1</h3>

5x + 3 is positive, then the inequality becomes:

  • 5x + 3 < 2
  • 5x < 2 - 3
  • 5x < - 1
  • x < - 1/5 or x < - 0.2

<h3>Option 2</h3>

5x + 3 is negative, then the inequality becomes:

  • - (5x + 3) < 2
  • 5x + 3 > - 2
  • 5x > - 2 - 3
  • 5x > - 5
  • x > - 1

The solution is the combination of the two intervals:

  • x > - 1 and x < - 0.2 ⇒ - 1 < x < - 0.2

Correct choice is B

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3 years ago
A) Let X be a random variable that can assume only positive integer values, and assume its probability function is P(X -n) A/3^n
kramer

Answer:

a) The value of A = 2

b) The value of B  = \dfrac{1}{12}

Step-by-step explanation:

a)

Given that:

X should be the random variable that assumes only positive integer values.

The probability function; P[X = n] = \dfrac{A}{3^n} for some constant A and n ≥ 1.

Then, let \sum \limits ^{\infty}_{n =1} P[X =n] = 1

This implies that:

A \sum \limits ^{\infty}_{n =1} \dfrac{1}{3^n}= 1

A \times  \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = 1

A \times  \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} = 1

A \times \dfrac{1}{2}=1

A = 2

Thus, the value of A = 2

b)

Suppose X represents a e constant A (n> 1). Find A.

b) Let X be a continuous random variable that can assume values between 0 and 3

Then, the density function of x is:

f_x(x) = \left \{ {{B(x^2+1)}   \ \ \ 0 \le x \le 3  \ \ \ \atop {0} \ \ \ otherwise} \right.

where; B is constant.

Then, using the property of the probability density function:

\int ^3_0 \ B (x^2+1 ) \ dx = 1\\

Taking the integral, we have:

B \Big [\dfrac{x^3}{3} +x \Big ]^3_0 = 1

B \Big [\dfrac{3^3}{3} +3 \Big ]= 1

B \Big [\dfrac{27}{3} +3 \Big ] = 1

B [ 9 +3 ] = 1

B [ 12 ] = 1

Divide both sides by 12

B  = \dfrac{1}{12}

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3 years ago
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