If a = b + 6 but a also = b * 3, what are the variables?

Mathematics

2 answers:

Ratling [72]2 years ago

4 0

The b would equal 3 because 3 + 6 = 9 and 3 * 3 = 9

netineya [11]2 years ago

4 0

Answer:

a = 9, b = 3

Step-by-step explanation:

Given:

a = b + 6
a = b*3

Then, eliminate a:

b + 6 = 3b
2b = 6
b = 3

Find a:

a = 3+ 6 = 9

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When you have two shapes to compare on a coordinate plane, you can determine the scale factor, knowing that the transformation w

musickatia [10]

Assume that the initial coordinates are (x,y) and that the dilated coordinates are (x',y').

The dilation is therefore:
(x,y) ............> (x',y')

Now, let's assume that the dilation factor is k.
Therefore:
x' = kx
y' = ky

Based on the above, all the student has to do is get the initial coordinates and the final ones and then substitute in any of the above two equations to get the value of k.

Example:
Assume an original point at (2,4) is dilated to coordinates (4,8). Find the dilation factor.
Assume the dilation coefficient is k.
(x,y) are (2,4) and (x',y') are (4,8)
Therefore:
x' = kx .........> 4 = k*2 ..........> k = 2
or:
y' = ky ..........> 8 = k*4 .........> k = 2
Based on the above, the dilation coefficient would be 2.

Hope this helps :)

8 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

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