The Aon Centre in Chicago,Illinois is one of the tallest buildings.built in1973

Mathematics

1 answer:

Dmitry [639]2 years ago

5 0

Yep
that's correct I checked on a website which and the question is kinda wired

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Which of the following equations has infinitely many solutions?

Valentin [98]

Answer:

Step-by-step explanation:

If you look at A and you subtract 3x by 4x you get x and you subtract 3 from -7 and you get -10. But if you look at B you do the stuff in the parentheses first and you get 4x+12x+28=16x+28 and they're the same so it's B.

7 0

2 years ago

Syracuse and New York City are 240 miles apart. One car left Syracuse heading for New York City at an average of 65 mph. A secon

bazaltina [42]

Let x be the distance from Syracuse where they pass. The first car travels a distance of x in time x/65 while the second car travels a distance 240-x in time (240-x)/55. They pass at the same time after leaving their starting points so x/65=(240-x)/55.
Cross-multiplying we get: 55x=65(240-x)=15600-65x, 120x=15600, x=15600/120=130 miles.
They pass 130 miles from Syracuse.

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2 years ago

PLEASE HELP?:((((((((

harkovskaia [24]

Answer:

Step-by-step explanation:

4 0

2 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

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