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3 years ago

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

You might be interested in

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

3 0

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

Hope this Helps!!!

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!

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

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

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.

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

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