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2 years ago

PLS PLS help!

Mathematics

1 answer:

Leto [7]2 years ago

4 0

9514 1404 393

Answer:

y = 3x^2 +30x +69

Step-by-step explanation:

Transformations work this way:

g(x) = k·f(x) . . . . vertical stretch by a factor of k

g(x) = f(x -h) +k . . . . translation (right, up) by (h, k)

So, the translation down 2 units will make the function be ...

f(x) = x^2 ⇒ f1(x) = f(x) -2 = x^2 -2

The vertical stretch by a factor of 3 will make the function be ...

f1(x) = x^2 -2 ⇒ 3·f1(x) = f2(x) = 3(x^2 -2)

The horizontal translation left 5 units will make the function be ...

f2(x) = 3(x^2 -2) ⇒ f2(x +5) = f3(x) = 3((x +5)^2 -2)

The transformed function equation can be written ...

y = 3((x +5)^2 -2) = 3(x^2 +10x +25 -2)

y = 3x^2 +30x +69

The attachment shows the original function and the various transformations. Note that the final function is translated down 6 units from the original. That is because the down translation came <em>before</em> the vertical scaling.

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

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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

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

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