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3 years ago

Is 1/2(12x + 24) and -6 (-x- 2)

Mathematics

1 answer:

ikadub [295]3 years ago

6 0

I hope this helps you

12x/2+24/2

6x+12

-6 ( -x-2)

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Determine whether the statement describes an algebraic expression or an algebraic equation.

ankoles [38]

Answer:

a) Algebraic equation

b) Algebraic equation

c) Algebraic expression

d)Algebraic equation

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3 years ago

Fill in the blanks below to make a true statement. In a binomial experiment with n trials and probability of success p, if np (1

Sergeu [11.5K]

Answer:

$\mu_{x}=np$

$\sigma^2_{X}=np(1-p)$

Step-by-step explanation:

∵ When x is a random variable having distribution B(n, p), then for sufficiently large value of n, the following random variable has a standard normal distribution,

$\frac{x-\mu}{\sigma}\sim N(0,1)$

Where,

$\mu=np$

$\sigma^2=np(1-p)$

Here the variable X has a binomial distribution,

Such that, np (1 - p) ≥ 10 ⇒ n is sufficiently large.

Where, n is the total numbers of trials, p is success in each trials,

So, the mean of variable X is,

$\mu_{X} = np$

And, variance of variable X is,

$\sigma^2{X}=np(1-p)$

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3 years ago

draw a rectangle give the length and the width to show that it has an area of 12cm make sure one side is longer than the other t

Elden [556K]

To find area, you multiply length * width. There are 4 sides on a number, so we need to have a equation like this. x * x * y * y = 12cm. If we do 4 * 4, we get 8. We could also use 2 * 2 = 4. 8 + 4 = 12. So, width = 8, and length = 4.

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4 years ago

MATH HELP!!! 100PTS!!!

Lostsunrise [7]

Answer:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Step-by-step explanation:

Given functions:

$p(x)=2x-4$

$r(x)=\dfrac{6x-1}{9x+1}$

Solve for p(x) = r(x):

$\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}$

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=18, \quad b=-40, \quad c=-3$

Substitute the values of a, b and c into the quadratic formula and solve for x:

$\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}$