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3 years ago

How is 1/2 + 1/4 solved differently than 1/2 + 1/3

Mathematics

1 answer:

ehidna [41]3 years ago

6 0

Look at the pictures attached.

Hope it helps! :)

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Least to greatest 6 3/10,1 3/10, 1 3/11, -7 6/14, 5 3/10

Drupady [299]

Answer:

-7 6/14, 1 3/11, 1 3/10, 5 3/10, 6 3/10

3 0

3 years ago

Please explain your answer

Irina18 [472]

$\text{Distance = } \sqrt{(5 + 8)^2+ (-3 + 3)^2 }= \sqrt{13^2} = 13$

Answer: 13 units

6 0

2 years ago

Help me please I beg if you can

Virty [35]

Answer: The mean is 3780

Step-by-step explanation:

Mean is found by adding all values and dividing the sum by how many values were added.

4250+4019+3895+3739+3401+3376=22680
22680/6=3780

6 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

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

7 0

3 years ago

Write 4670 000 in standard form.

lianna [129]

Answer:

4.67×10^6 is the correct answer

7 0

2 years ago

Read 2 more answers