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rjkz [21]
3 years ago
9

Transform the equation to isolate x: ax = bx + 1. How is

Mathematics
2 answers:
shepuryov [24]3 years ago
6 0

Answer:

x=\frac{1}{a-b}

The value of x is inversely proportional to the difference of <em>a</em> and <em>b</em>.

Step-by-step explanation:

The given expression is ax-bx+1

We factor the left hand side to get:

(a-b)x=1

We divide both sides by: a-b

x=\frac{1}{a-b}

Observe that the difference of a and b is in the denominator.

The whole is of the form: x\propto \frac{1}{a-b}

The value of x is inversely proportional to the difference of <em>a</em> and <em>b</em>.

andrey2020 [161]3 years ago
5 0

Answer:

Sample Response: <em>The equation ax = bx + 1 is the same as x = 1/(a  - b) when solved for x. This means that x is equal to the reciprocal of the difference of a and b. </em>

this is correct on edge

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3 3/5 x 1/4 *Simply your answer if you can
3241004551 [841]

Answer:

9/10 or 0.9

Step-by-step explanation: hope this helps

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3 years ago
Based on a​ poll, among adults who regret getting​ tattoos, 1212​% say that they were too young when they got their tattoos. Ass
photoshop1234 [79]

Answer:

0.3165

Step-by-step explanation:

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of adults sampled = 9

x = Number of successes required = number of selected adults that say they were too young to get tattoos = 0

p = probability of success = probability of an adult that regrets getting a tattoo saying they were too young to get the tattoo = 12% = 0.12

q = probability of failure = 1 - p = 1 - 0.12 = 0.88

P(X=0) = ⁹C₀ (0.12)⁰ (0.88)⁹⁻⁰

P(X=0) = 0.31647838183 = 0.3165

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7 0
3 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

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tekilochka [14]

Answer:

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Step-by-step explanation:

fbgetjmnryjnmyjnmryjnhm

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