What is the solution (a, b) to this system of linear equations? 3a+6b=45 2a-2b=-12

WILL GIVE BRAINLIEST

suter [353]

Answer:

$\sqrt{65}$ is the closest to 8

Step-by-step explanation:

$\sqrt{61}=7.81 \\\sqrt{66}=8.12 \\\sqrt{65} =8.06\\\sqrt{62} =7.87$

4 0

2 years ago

17. Find the ratio of the perimeter to the area of the square shown. x + 3 x + 3

Serga [27]

Answer:

4x+12/x^2+9

Step-by-step explanation:

Perimeter

x+3+x+3+x+3+x+3

Area

(x+3)(x+3)

7 0

2 years ago

I will award you brainlist

ExtremeBDS [4]

Answer:

TRUE

Step-by-step explanation:

We are given 2 triangles, ∆ABC and ∆DEF.

For the two trinagles to be considered similar, both must have their set of corresponding angles congruent to each other. That is, their corresponding angles are equal.

From the information given, the following are the set of corresponding angles:

<B = <E = 31°

<B = <D = 90°

<C = <F = 59° [180 - (90+31)]

The corresponding angles of both triangles are congruent. Therefore, it is guaranteed that ∆ABC is similar to ∆DEF (∆ABC ~ DEF).

3 0

3 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

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