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3 years ago

4x + 7 = 5 solve for x

Mathematics

2 answers:

meriva3 years ago

7 0

Answer:

-1/2

Step-by-step explanation:

4x + 7 = 5

4x=5-7

4x=-2

x=-2/4

x=-1/2

hope this helps u!

amid [387]3 years ago

7 0

Answer:

0.5

4x+7=5

4x=5-7

4×=-2

4\=/4

answer is -0.5

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F(x)=bx^2+32 For the function f defined above, b is a constant and f(2)=40. What is the value of f(-2)?

zhuklara [117]

Answer:

f(-2) = 0

Step-by-step explanation:

Given that:

f(x) = bx^2 + 32

f(2) = b(2)^2 + 32

f(2) = 4b + 32

f(2) = 4b = -32

f(2) = b = -32/4

f(2) = b = -8

Thus;

f(-2) = -8(-2)^2 + 32

f(-2) = -8(4) + 32

f(-2) = -32 + 32

f(-2) = 0

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Write 14/25 as a decimal. Show work

Fudgin [204]

Answer:

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8 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

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