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3 years ago

What is the decimal representation of 5/11?

Mathematics

2 answers:

stepan [7]3 years ago

5 0

The answer to this question is 45

miv72 [106K]3 years ago

5 0

Simple my dear sir, divide 5 by 11, and BAM .45 (infinite) so its .45 with a line over it

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F(x) = -x^3+ 6x – 7 at x = 2

vodomira [7]

Answer:

f(2)= -4

Step-by-step explanation:

f(x)= -x^3 + 6x - 7

f(2) = -2^3 + 6*2 - 7

f(2)= -8 +12 -7

f(2)= 3-7

f(2)= -4

4 0

2 years ago

It is a well-known fact that 50% of the general population are rascals (P(R) = 0.5) and 50% of the general population are not ra

aleksley [76]

Answer:

The answer is D. 0.75

Step-by-step explanation:

Let call R the event that a person is rascal, RC a person is not rascal, TH a person use top hat and NTH a person don’t use top hat.

From the information on the question we have 4 options with their respective probability:

1. A person could be rascal and use top Hat: this probability is calculate as the multiplication of the probability of be a rascal (0.5) and use a top hat (0.9), then:

P(R y TH)=0.5*0.9=0.45

2. A person could be rascal and don’t use top Hat: this probability is calculate as the multiplication of the probability of be a rascal (0.5) and not use a top hat (0.9), then:

P(R y NTH)=0.5*0.1=0.05

3. A person could be not rascal and use top Hat: this probability is calculate as the multiplication of the probability of not be a rascal (0.5) and use a top hat (0.3), then:

P(RC y TH)=0.5*0.3=0.15

4. A person could be not rascal and not use top Hat: this probability is calculate as the multiplication of the probability of not be a rascal (0.5) and not use a top hat (0.7), then:

P(RC y NTH)=0.5*0.7=0.35

Then the probability that a person is a rascal given that he is wearing a top hat could be written and calculate as:

$P(R/TH)=\frac{P(R y TH)}{P(TH)}$

For calculate P(TH) we need to sum all the option in which TH is involve so:

P(TH) = P(R y TH)+ P(RC y TH)

P(TH)=0.45+0.15=0.6

Replacing values on the first equation we get:

$P(R/TH)=\frac{0.45}{0.6} =0.75$

So, the probability that a person is a rascal given that he is wearing a top hat is 0.75

5 0

3 years ago

I guess I'm lacking in differential equations. I couldn't solve this question. Can you help me?

Sonja [21]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Reciprocals

Algebra II

Log/Ln Property: $ln(\frac{a}{b} ) = ln(a) - ln(b)$

Calculus

Derivatives

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

Derivative of Ln: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

$ln(\frac{2x-1}{x-1} )=t$

Step 2: Differentiate

Rewrite: $t = ln(\frac{2x-1}{x-1})$
Rewrite [Ln Properties]: $t = ln(2x-1) - ln(x - 1)$
Differentiate [Ln/Chain Rule/Basic Power Rule]: $\frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1$
Simplify: $\frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}$
Rewrite: $\frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}$
Combine: $\frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}$
Reciprocate: $\frac{dx}{dt} = -(2x-1)(x-1)$
Distribute: $\frac{dx}{dt} = (1-2x)(x-1)$