<img src="https://tex.z-dn.net/?f=%5Cbf%205x-6%3D3x-8" id="TexFormula1" title="\bf 5x-6=3x-8" alt="\bf 5x-6=3x-8" align="absmidd

All categories

Helen [10]

1 year ago

le" class="latex-formula">

Mathematics

2 answers:

Vaselesa [24]1 year ago

5 0

Subtract 3x from 5x you get 2x the Add 6 to -8 which is -2 then divide 2 by -2 and get X=-1

natulia [17]1 year ago

4 0

X = -1

I’d explain but it’s long to type up and i cba bothering lol

You might be interested in

There is a game at the carnival that costs $20 to play. There are two outcomes: either you win $100 , or you lose. To play you r

Taya2010 [7]

Outcome on winning = $100 - $20 = $80
Outcome on losing = $20

Probability of winning = 1/6
Probability of losing = 5/6

Expected Value = 1/6 (80) +5/6(-20) = -3.333

This shows on average from each game, the game earns $ 3.333.

So, if 1000 such games are played, the game will earn 1000 x 3.33 = $3330

So, the answer to this question is option A

5 0

2 years ago

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

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