5. Find the value of x. A. 55 B. 21 C. 12 D. 22 3x -30

Find the value ofx. O is the center of the circle. Round your answer to the nearest hundredth.

lyudmila [28]

Answer:

A. In a binomial distribution, the value ofx represents the number of successes in n trials, while in a geometric distribution, the value ofx represents the first trial that results in a success.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

The most important difference is that in the binomial distribution, the value of x represents the successes in n trials.

And by the other hand in the geometric distribution, x represents the number of failures before you get a success in a series of Bernoulli trials.

So then the best answer for this case is:

A. In a binomial distribution, the value of x represents the number of successes in n trials, while in a geometric distribution, the value ofx represents the first trial that results in a success.

Step-by-step explanation:

8 0

3 years ago

Solve by graphing.

const2013 [10]

Answer:

(4,4)

Step-by-step explanation:

y = 2x - 4 -------eqn 1

4y = x + 12 ------eqn 2

From eqn 1 y = 2x - 4; insert in eqn 2

4 (2x-4) = x + 12 (open the bracket)

8x - 16 = x + 12 (taking like figures to the same side, sign changes)

8x - x = 12 + 16

7x = 28

x = 28/7

x = 4

From eqn 1

y = 2x - 4

y = 2(4) - 4

y = 8 - 4

y = 4

(x,y) = (4,4)

8 0

3 years ago

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

2 years ago

Find the values of x and y.

Vesna [10]

Answer:

1500

Step-by-step explanation:

you multiply 30x50

6 0

3 years ago

Why does the tail approaches the x-axis but never touches it

Brut [27]

Because most of the area under any normal curve falls within a limited range of the number line

7 0

2 years ago