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Helen [10]
3 years ago
7

3-3x6+2 and how is that reached, please!

Mathematics
1 answer:
Dvinal [7]3 years ago
3 0
The order of mathematical operation is MDAS.
M - multiplication
D - division
A - addition
S - subtraction

3 - 3 x 6 + 2

Multiplication : 3 x 6 = 18
Division: none
Addition : + 3 + 2 = 5 (combine positive numbers)
Subtraction : 5 - 18 = -13

3 - 18 + 2 = -13
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Hello again! This is another Calculus question to be explained.
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Answer:

See explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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  • Left to Right

<u>Algebra I</u>

Functions

  • Function Notation
  • Exponential Property [Rewrite]:                                                                   \displaystyle b^{-m} = \frac{1}{b^m}
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<u>Calculus</u>

Differentiation

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  • 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:

  1. f(x) = cxⁿ
  2. 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 <em>x</em> = 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 <em>x</em> = 2, simply substitute in <em>x</em> = 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

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Answer:

"0.0125" is the right solution.

Step-by-step explanation:

The given values are:

Random sample,

n = 90

Claims,

p = 20%

or,

  = 0.20

By using normal approximation, we get

⇒  \mu = np

On substituting the values, we get

⇒      =90\times 0.20

⇒      =18

Now,

The standard deviation will be:

⇒  \sigma=\sqrt{np(1-p)}

On putting the above given values, we get

⇒      =\sqrt{90\times 0.20\times (1-0.20)}

⇒      =\sqrt{18\times 0.8}

⇒      =\sqrt{14.4}

⇒      =3.7947

hence,

By using the continuity correction or the z-table, we get

⇒  P(x < 10) = P(x < 9.5)

⇒  P(x < 10) = P(\frac{x-\mu}{\sigma} -\frac{9.5-18}{3.7947} )

⇒  P(x < 10) = P(Z < -2.24)

From table,

⇒  P(x < 10) = 0.0125

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3 years ago
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