All categories

3 years ago

(81m^6) 1/2 (Shift +6)

Mathematics

1 answer:

Katarina [22]3 years ago

6 0

Step-by-step explanation:

$\sqrt{81 {m}^{6} }$

= 9m³

You might be interested in

Could someone please help me with this thank you very much

elena-s [515]

Answer:

Answer for #10:

$y = 2 \: and \: x = 1$

Answer for #1:

$x = 2 \: and \: y = - 1$

Step-by-step explanation:

In order to solve a simultaneous equation using the substitution method you have to:

Use one of the equation and substitute it into the other equation, in this case equation B was substituted into equation A.

 $equ. \: a = (2x - 3y = 7) \\ equ. \: b = (x = 1 - y)$ 

Therefore since x = 1 - y, equation A would be:

$2(1 - y) - 3y = 7 \\ (2 - 2y) - 3y = 7 \\ - 2y - 3y = 7 - 2 \\ - 5y = 5 \\ y = - 1 \: and \: x = 2$

8 0

1 year ago

The endpoints of ab are a(1,4) and b(6,-1). If point c divides ab in the ratio 2:3, the coordinates of c are ?. If point d divid

aliya0001 [1]

Answer:

see explanation

Step-by-step explanation:

Find the coordinates of c using the section formula

$x_{c}$ = $\frac{(3(1)+(2(6)}{2+3}$ = $\frac{3+12}{5}$ = 3

$y_{c}$ = $\frac{(3(4))+(2(-1))}{2+3}$ = $\frac{12-2}{5}$ = 2

coordinates of c = (3, 2)

Similarly to find the coordinates of d

$x_{d}$ = $\frac{x(2(1))+(3(3))}{3+2}$ = $\frac{2+9}{5}$ = $\frac{11}{5}$

$y_{d}$ = $\frac{(2(4))+(3(2))}{3+2}$ = $\frac{8+6}{5}$ = $\frac{14}{5}$

coordinates of d = ( $\frac{11}{5}$ , $\frac{14}{5}$ )

4 0

2 years ago

Read 2 more answers

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

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