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3 years ago

The general equation for an ellipse?

Mathematics

1 answer:

allochka39001 [22]3 years ago

6 0

Answer:

The standard equation for an ellipse, x 2 / a 2 + y2 / b 2 = 1

Step-by-step explanation:

it represents an ellipse centered at the origin and with axes lying along the coordinate axes.

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What is symmetry? super easy i know

AVprozaik [17]

Answer:

i'm confused if you know why are you asking

Step-by-step explanation:

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3 years ago

You are dealt one card from a 52-card deck. Find the probability that you are not dealt a club

saveliy_v [14]

More card math! Easy!!

Exactly 1/4 of a 52-card deck of cards is clubs. Another 1/4 spades, another 1/4 is hearts, and the final 1/4 is diamonds. To find what one fourth of 52is, you can divide 52 by 4. That equals 13. So there are 13 clubs in the entire deck of cards. If you are trying <em>not </em>to get a club, you have to find the remainder of the cards without those 13. 52 - 13 = 39. This means that you have 39/52 odds of <em>not </em>getting a club. Simplified you get 3/4.

I did it the long way. A shorter was is to see that, obviously, one whole minus one fourth is three fourths.

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3 years ago

Rewrite 18y x 4x using Commutative Property

Vsevolod [243]

The answer to the question

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3 years ago

Point M (-3, 6) and point 'N (1,-1) are located on a coordinate

lisabon 2012 [21]

The distance between point M and point is is about 8.06 units (rounded to nearest hundredth)

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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

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

7 0

3 years ago