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3 years ago

6

aey sample consisting of a portion of the population that is not scientifically drawn but instead collected because it easy to a

ccess is called

1 answer:

hram777 [196]3 years ago

7 0

Answer:

See below

Step-by-step explanation:

A sample consisting of a portion of the population that is not scientifically drawn but instead collected because it easy to access is called ...

accidental sample
or grab sample
or convenience sample
or opportunity sample

All above refer to same method

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Need help please. I don't think this is dimensionally correct however I want to be sure. Please check if s = (2)(u+v)t^2 is dime

irga5000 [103]

Your equation is not dimensionally consistent.

Let L = length and T = time. Without taking into account precise units, you can write velocity as the ratio between some distance L and some time T.

So in terms of these fundamental quantities, your equation can be written as

L = (L/T + L/T)T^2

L = (L + L)T

L = LT

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4 years ago

How do you solve the function 5x-4=-2(3x+2)

NemiM [27]

5x-4=-2(3x+2)
5x-4= -6x-4
5x+6x= -4+4
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4 years ago

Read 2 more answers

According to the diagram, point B lies on how many faces of the pyramid?

Scrat [10]

Answer:

POINT B LIES ON FOUR FACES OF THE PYRAMID

Step-by-step explanation:

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3 years ago

Hi, how do we do this question?

Nutka1998 [239]

Answer:

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients
Factoring

<u>Algebra II</u>

Polynomial Long Division

<u>Calculus</u>

Differentiation

Derivatives
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:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
Integration Constant C
Indefinite Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

[Integrand] Rewrite [Polynomial Long Division (See Attachment)]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$
[1st Integral] Reverse Power Rule: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$

<u>Step 3: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 3x + 1$
[<em>u</em>] Differentiate [Basic Power Rule]: $\displaystyle du = 3 \ dx$

<u>Step 4: Integrate Pt. 3</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx$
[Integral] U-Substitution: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C$
Back-Substitute: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C$
Factor: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3} \bigg) + C$
Rewrite: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

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

Unit: Integration

Book: College Calculus 10e

8 0

3 years ago

Ayuda es urgenteeeee

Luda [366]

Answer:

ok cual es la pregunta?

Step-by-step explanation:

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3 years ago