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Minchanka [31]
4 years ago
7

Based on a survey, assume that 28% of consumers are comfortable having drones deliver their purchases. Suppose that we want to f

ind the probability that when consumers are randomly selected, exactly of them are comfortable with delivery by drones. Identify the values of n, x, p, and q. 28
Mathematics
1 answer:
borishaifa [10]4 years ago
6 0

<em>Question:</em>

<em>Based on a survey, assume that 28% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones. </em>

<em>Identify the values of n, x, p, and q.</em>

Answer:

n = 5

x = 3

p = 0.28

q = 0.72

Step-by-step explanation:

Given

Proportion of customers that prefer drone = 28%

Selected customers= 5

Customers expected to prefer drone = 3

Required

Determine the value of n, x, p and q

In probability; n represents the selected sample;

So;

n = 5

The expected number of people to prefer drone is represented by x;

So;

x = 3

In probability, opposite probabilities add up to 1

i.e.

p + q = 1

Where p is the given probability;

p = 28\%

Convert to decimal

p = 0.28

Substitute 0,28 for p in the above equation

p + q = 1 becomes

0.28 + q = 1

Make q the subject of formula

q = 1 - 0.28

q = 0.72

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Sonja uses 8 of the 12 eggs in her dozen eggs to make a breakfast casserole. Olga uses 5/6 of a dozen eggs to make scrambled egg
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3 years ago
Harvey the hamster can run 3 and 1/6 km in 1/4 of an hour. What is his average speed in kilometers per hour?
liubo4ka [24]
<h3>Answer:  12 & 2/3 km per hour</h3>

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As an improper fraction, 12 & 2/3 = 38/3

==================================================

Explanation:

Let's use this formula

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3 & 1/6 = (3*6+1)/6

3 & 1/6 = 19/6

The mixed number 3 & 1/6 is the same as the improper fraction 19/6

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We can express that as the ratio

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3 years ago
The third-degree Taylor polynomial about x = 0 of In(1 - x) is
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Answer:

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

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Functions
  • Function Notation

<u>Calculus</u>

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.

<u />

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = ln(1 - x)

Center: x = 0

<em>n</em> = 3

<u>Step 2: Differentiate</u>

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

<u>Step 3: Evaluate Functions</u>

  1. Substitute in center <em>x</em> [Function]:                                                                     \displaystyle f(0) = ln(1 - 0)
  2. Simplify:                                                                                                             \displaystyle f(0) = 0
  3. Substitute in center <em>x</em> [1st Derivative]:                                                             \displaystyle f'(0) = \frac{1}{0 - 1}
  4. Simplify:                                                                                                             \displaystyle f'(0) = -1
  5. Substitute in center <em>x</em> [2nd Derivative]:                                                           \displaystyle f''(0) = \frac{-1}{(0 - 1)^2}
  6. Simplify:                                                                                                             \displaystyle f''(0) = -1
  7. Substitute in center <em>x</em> [3rd Derivative]:                                                            \displaystyle f'''(0) = \frac{2}{(0 - 1)^3}
  8. Simplify:                                                                                                             \displaystyle f'''(0) = -2

<u>Step 4: Write Taylor Polynomial</u>

  1. 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
  2. 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

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Varvara68 [4.7K]
<span>8.42 x 10^3
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8420 is between 
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