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3 years ago

L

Mathematics

2 answers:

Brums [2.3K]3 years ago

8 0

Answer:

$539.00

Step-by-step explanation:

K=98

El=3K 3*98=294

Em=.5El 294*.5=147

K+El+Em=98+294+147=539

egoroff_w [7]3 years ago

5 0

Answer:

$539

Step-by-step explanation:

First we need to look at the information given

Karen: $98.00

Emily: Karen x 3

Elizabeth: Emily /2

So if karen donated $98 and emily donated 3 times that all we need to do is multiply it by 3

$98*3= $294

So we now know that Emily donated $294

If Emily donated $294 and elizabeth donated half that all we need to do is divide $294 by 2

$294/2= $147

So we now know that Elizabeth donated $147

All we need to do know is add all three answers together $98+$294+$147=

$539

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Naomi solves a system of equations using substitution. The system has infinitely many solutions. Which of the following could be

Sedbober [7]

Answer:

Option C) 2=2 is the last step of Naomi and it is the correct answer

Step-by-step explanation:

Given that Naomi solves a system of equation by Substitution method :

The given system has infinite many solutions

Naomi's last step is 2=2

<h3>For : A system of equations after solved then its constant value must be same in both the sides</h3>

Then the system of equations has infinite many solutions

Therefore it must be 2=2

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Answer:

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Step-by-step explanation:

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3 years ago

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Umnica [9.8K]

<span>2(8x - 4) = 32 distribute through parenthesis
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3 years ago

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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

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

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3 years ago