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2 years ago

Please answer this I will give brainliest if you answer right first.

Mathematics

1 answer:

Vesnalui [34]2 years ago

3 0

Answer:

Step-by-step explanation:

The constant of proportionality is the value of y/x. Picking a random value on the graph, such as (2, 10), we can substitute this into the formula for k = y/x. 10/2 = 5. This satisfies one portion of the answer. Now, if we take the value of x from the same point, (2, 10), which is 2, and multiply it by 5, as illustrated in y = 5x, we get 10 for y. This suits the equation as mentioned and we can check other points with the same process to ensure that this answer is correct.

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Answer:

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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

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

7 0

3 years ago

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katrin [286]

Answer:

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2 years ago

Which equation has a graph that is a parabola with a vertex at (5,3)?

lara [203]

Answer:

y = (x -5)² + 3.

Step-by-step explanation:

Given : parabola with a vertex at (5,3).

To find : Which equation has a graph that is a parabola.

Solution : We have given vertex at (5,3).

Vertex form of parabola : y = (x -h)² + k .

Where, (h ,k ) vertex .

Plug h = 5 , k= 3 in vertex form of parabola.

Equation :y = (x -5)² + 3.

Therefore, y = (x -5)² + 3.

6 0

3 years ago