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Evgesh-ka [11]
3 years ago
8

What is the simplified form of 3abc divided by abc when abc does not equal zero

Mathematics
1 answer:
melisa1 [442]3 years ago
4 0
3abc/abc = 3.
Simply put, the answer if you divide this equation by ABC, will quite literally transform into 3.

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What is the answer to 9x(6+x)
Anna11 [10]

54x + x \\  \\ (constributive \: property)
5 0
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
You want to solve 11% of what number is 22. Explain how you can solve that by using a proportion
Doss [256]
Write the first ratio as 11/100 and set it equal to 22/n.

since 11 x 2=22 you can multiply 100 x 2 and you'll get 200. 

*Check*: 11% of 200=22
3 0
3 years ago
Given the terms a10 = 3 / 512 and a15 = 3 / 16384 of a geometric sequence, find the exact value of the term a30 of the sequence.
GenaCL600 [577]
The answer is \frac {3}{2^{29}}
5 0
3 years ago
What are the solutions to the system of equations? y=2x^2−6x+3 y=2x+3
adoni [48]

Answer:

Step-by-step explanation:

y=2x²-6x+3

y=2x+3

2x²-6x+3=2x+3

2x²-6x+3-2x-3=0

2x²-8x=0

2x(x-4)=0

x=0,4

y=0+3=3

and y=2(4)+3=11

so solutions are (0,3) and (4,11)

5 0
2 years ago
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