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3 years ago

To find the quotient of 8 :- 1/3, multiply 8 by

Mathematics

2 answers:

MrMuchimi3 years ago

7 0

Answer:

Step-by-step explanation:

Elina [12.6K]3 years ago

4 0

Answer:

To find the quotient of 8 Divided by 1/3.

Step-by-step explanation

Ur answer would 3 but to find out if its right is...

U would do 8 divided by 1/3 multiply 8 by three your answer would be 1/3 or 3

Hope It Helps (:

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According to the Department of Transportation, 27% of domestic flights were delayed in the last year at JFK airport. Five flight

Mars2501 [29]

Answer:

The probability that all the five flights are delayed is 0.2073.

Step-by-step explanation:

Let X = number of domestic flights delayed at JFK airport.

The probability of a domestic flight being delayed at the JFK airport is, P (X) = p = 0.27.

A random sample of n = 5 flights are selected at JFK airport.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={5\choose x}0.27^{x}(1-0.27)^{5-x};\ x=0,1,2...$

Compute the probability that all the five flights are delayed as follows:

$P(X=5)={5\choose 5}0.27^{5}(1-0.27)^{5-5}=1\times 1\times 0.207307=0.2073$

Thus, the probability that all the five flights are delayed is 0.2073.

4 0

3 years ago

Evaluate the following trigonometric expression

Harlamova29_29 [7]

Answer:

-60

hope it helps

5 0

3 years ago

5 pounds of tomatoes cost $15.50. How much does one pound of tomatoes cost?

Tanzania [10]

Answer:

C 3.10

Step-by-step explanation:

15.50 divide by 5 is 3.10

Brainliest?

3 0

3 years ago

Read 2 more answers

(cos6x+6cos4x+15cos2x+10)/(cos5x+5cos3x+10cosx)=?

Tasya [4]

Do you mean cos6x+6cos4x+15cos2x+10cos5x+5cos3x+10cosx ? or cos6x+6cos4x+15cos2x+10cos5x+5cos3x+10cosx ? or cos6x+6cos4x+15cos2x+10cos5x+5cos3x+10cosx ? or cos6x+6cos4x+15cos2x+10cos5x+5cos3x+10cosx ?

7 0

3 years ago

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