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3 years ago

Someone refresh me on constant of proportionality, forgot all about it!!!!

Mathematics

1 answer:

MAXImum [283]3 years ago

3 0

A cycle, that is in the consideration of the earth and also help the combinations of things

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Which equation represents a circle with a center at (–3, –5) and a radius of 6 units?

dolphi86 [110]

(x+3)²+(x+5)²=6²
(x+3)²+(x+5)²=36

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3 years ago

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dezoksy [38]

I don’t know spanish sorry

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3 years ago

The interest rate r required to increase your investment p to the amount a in t years is found by mc075-1.jpg. Find the interest

kotegsom [21]

I hope this is the answer :)
45.67=r

6 0

2 years ago

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Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Find an equation for the inverse of the relation y=6x-3. Show how you found the inverse equation.

Elis [28]

Hi there! The answer is y = 1/6x + 1/2

To find the inverse function x and y need to switch places.

$y = 6x - 3 \\ x = 6y - 3$

Now rewrite the equation, until it will be in standard form (slope-intercept form) again.
$x = 6y - 3$

Add 3.
$x + 3 = 6y$

Switch sides.
$6y = x + 3$

Divide both sides by 6.
$y = \frac{1}{6} x+ \frac{3}{6} \\ y = \frac{1}{6} x + \frac{1}{2}$

Hence, the answer is y = 1/6x + 1/2.
~ Hope this helps you!

4 0

2 years ago