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3 years ago

Fundamental Counting Principle (L2)

Mathematics

1 answer:

forsale [732]3 years ago

7 0

Answer:

As we dive deeper into more complex probability problems, you may start wondering, "How can I figure out the total number of outcomes, also known as the sample space?"

Step-by-step explanation:

hop this helps

You might be interested in

The curve

kherson [118]

Answer:

Point N(4, 1)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

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:

Step 1: Define

$\displaystyle y = \sqrt{x - 3}$

$\displaystyle y' = \frac{1}{2}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = (x - 3)^{\frac{1}{2}}$
Chain Rule: $\displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)$
Simplify: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1$
Multiply: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x - 3}}$

Step 3: Solve

Find coordinates

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle 1 = \frac{1}{\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply √(x - 3) on both sides: $\displaystyle \sqrt{x - 3} = 1$
[Equality Property] Square both sides: $\displaystyle x - 3 = 1$
[Addition Property of Equality] Add 3 on both sides: $\displaystyle x = 4$

y-coordinate

Substitute in x [Function]: $\displaystyle y = \sqrt{4 - 3}$
[√Radical] Subtract: $\displaystyle y = \sqrt{1}$
[√Radical] Evaluate: $\displaystyle y = 1$

∴ Coordinates of Point N is (4, 1).

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

Unit: Derivatives

Book: College Calculus 10e

4 0

2 years ago

the waiter places a bowl of soup in front of lacy in a counterclockwise direction she passes the soup 90° the person receiving t

dybincka [34]

haifia is the answer did it on test

5 0

3 years ago

In the given figure, identify which of the line segments is longer than XY. Question 15 options: A) TU B) PQ C) VW D) RS

dalvyx [7]

Answer:

A. T, U

Step-by-step explanation:

T and U are stretched across the paper. the others seem a bit close or smaller to the size of X and Y, but when you look at T and U they seem longer than X an Y

Hope this helps!!! :)

7 0

3 years ago

Write a real-world word problem that can be solved using elapsed time. Include the solution.

ElenaW [278]

Or...you could do this.

Lacey walks in a store at 8:30,the store opens at 12:00,how much time does Lacey have to get the store opened and stock the shelves?

4 1/2hours

3 0

3 years ago

Solve sin theta + 1 = cos 2 theta on the interval 0 less than or equal to theta <2 pi

hjlf

Answer:

Ф = 0 and Ф = π

Step-by-step explanation:

* Lets explain how to solve the problem

∵ sin Ф + 1 = cos²Ф, where 0 ≤ Ф < 2π

- To solve we must to replace cos²Ф by 1 - sin²Ф

∵ sin²Ф + cos²Ф = 1

- By subtracting sin²Ф from both sides

∴ cos²Ф = 1 - sin²Ф

- Lets replace cos²Ф in the equation above

∴ sin Ф + 1 = 1 - sin²Ф

- Subtract 1 from both sides

∴ sin Ф = - sin²Ф

- Add sin²Ф for both sides

∴ sin²Ф + sin Ф = 0

- Take sin Ф as a common factor from both sides

∴ sin Ф(sin Ф + 1) = 0

- Equate each factor by 0

∵ sin Ф = 0

∴ Ф = 0 OR Ф = 2π

∵ sin Ф + 1 = 0

- Subtract 1 from both sides

∴ sin Ф = -1

∴ Ф = π

∵ 0 ≤ Ф < 2π

∵ Ф < 2π

∴ We will refused the answer Ф = 2π

∴ Ф = 0 and Ф = π

7 0

3 years ago