All categories

3 years ago

What is the probability of choosing a

Mathematics

1 answer:

sladkih [1.3K]3 years ago

3 0

Answer:

Probability of choosing a penny at random is 1/5 or 0.2

Step-by-step explanation:

For many probabilities like this one, you have to add the total of materials or items together, even if they are different colors, shape, coin type, etc.

you have 12 pennies, 19 nickles, 14 dimes, and 15 quarters. This adds up to a total of 60 coins. Twelve of those are pennies.

You have twelve possible pennies to choose at random from a total of 60 in a bag.

So 12/60/ This simplifies to a probability of 1/5. If you divide 1 by 5 than it would be 0.2

The probability of choosing a penny from a bag of coins is 1/5, or 0.2.

Hope this helps

You might be interested in

Question 5 (Multiple Choice Worth 4 points)

loris [4]

(-2,2) (0,1)
slope = y1-y2/x1-x2 .... 2-1/-2-0 = -1/2 the answer is C

7 0

3 years ago

Read 2 more answers

PLESSE HELP ME ASAP

harina [27]

Answer:

t55=d

Step-by-step explanation:

55 is mph and since t is time t would be multiplied by 55

4 0

3 years ago

A cable is 60 decimeters long. How long is the cable in meters?

jonny [76]

60 decimeters to meters is 6 meters.

8 0

2 years ago

Read 2 more answers

The third-degree Taylor polynomial about x = 0 of In(1 - x) is

gizmo_the_mogwai [7]

Answer:

$\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

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

Calculus

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

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

MacLaurin/Taylor Polynomials

Approximating Transcendental and Elementary functions
MacLaurin Polynomial: $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n$
Taylor Polynomial: $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

Step 1: Define

Identify

f(x) = ln(1 - x)

Center: x = 0

n = 3

Step 2: Differentiate

[Function] 1st Derivative: $\displaystyle f'(x) = \frac{1}{x - 1}$
[Function] 2nd Derivative: $\displaystyle f''(x) = \frac{-1}{(x - 1)^2}$
[Function] 3rd Derivative: $\displaystyle f'''(x) = \frac{2}{(x - 1)^3}$

Step 3: Evaluate Functions

Substitute in center x [Function]: $\displaystyle f(0) = ln(1 - 0)$
Simplify: $\displaystyle f(0) = 0$
Substitute in center x [1st Derivative]: $\displaystyle f'(0) = \frac{1}{0 - 1}$
Simplify: $\displaystyle f'(0) = -1$
Substitute in center x [2nd Derivative]: $\displaystyle f''(0) = \frac{-1}{(0 - 1)^2}$
Simplify: $\displaystyle f''(0) = -1$
Substitute in center x [3rd Derivative]: $\displaystyle f'''(0) = \frac{2}{(0 - 1)^3}$
Simplify: $\displaystyle f'''(0) = -2$

Step 4: Write Taylor Polynomial

Substitute in derivative function values [MacLaurin Polynomial]: $\displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3$
Simplify: $\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e

5 0

3 years ago

The three cards below form a number pattern

Marrrta [24]

You do not have a picture inserted it is just a grey screen

3 0

3 years ago