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4 years ago

(-5/6) + c = (-11/12)

2 answers:

7 0

Remember you can do anything to an equation as long as you doit t both sides

-5/6+c=-11/12
add 5/6 to both sides
5/6-5/6+c=-11/12+5/6
0+c=-11/12+5/6
c=-11/12+5/6
how do we add?
we make same denomenator

11/12 and 5/6
6 times 2=12 so
5/6 times 2/2=10/12

c=-11/12+5/6
c=-11/12+10/12
c=-1/12

rosijanka [135]4 years ago

4 0

(-5/6) + c = (-11/12)
c=-11\12+5\6
c=19\12 or 1 7\12

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If f(x) and f^-1(X) are inverse functions of each other and f(x)=2x+5, what I f^-1(8)

podryga [215]

Answer: 3/5

Step-by-step explanation:

First, you need to find the inverse of f(x)=2x+5 .

$f-1(x)= \frac{x}{2} -\frac{5}{2}$

Then, you'd plug in 8 into that equation, which I got $\frac{3}{5\\}$

Hope this helps :)

3 0

3 years ago

Let Y1 and Y2 be independent exponentially distributed random variables, each with mean 7. Find P(Y1 > Y2 | Y1 < 2Y2). (En

ArbitrLikvidat [17]

Y₁ and Y₂ are independent, so their joint density is

$f_{Y_1,Y_2}(y_1,y_2)=f_{Y_1}(y_1)f_{Y_2}(y_2)=\begin{cases}\frac1{49}e^{-\frac{y_1+y_2}7}&\text{for }y_1\ge0,y_2\ge0\\0&\text{otherwise}\end{cases}$

By definition of conditional probability,

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = P((Y₁ > Y₂) and (Y₁ < 2 Y₂)) / P(Y₁ < 2 Y₂)

Use the joint density to compute the component probabilities:

• numerator:

$P((Y_1>Y_2)\text{ and }(Y_1$

$=\displaystyle\frac1{49}\int_0^\infty\int_{\frac{y_1}2}^{y_1}e^{-\frac{y_1+y_2}7}\,\mathrm dy_2\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\int_{-\frac{3y_1}{14}}^{-\frac{2y_1}7}e^u\,\mathrm du\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\left(e^{-\frac{2y_1}7} - e^{-\frac{3y_1}{14}}\right)\,\mathrm dy_1$

$=\displaystyle-\frac17\left(-\frac72e^{-\frac{2y_1}7} + \frac{14}3 e^{-\frac{3y_1}{14}}\right)\bigg|_0^\infty$

$=\displaystyle-\frac17\left(\frac72 - \frac{14}3\right)=\frac16$

• denominator:

$P(Y_1$

(I leave the details of the second integral to you)

Then you should end up with

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = (1/6) / (2/3) = 1/4

5 0

3 years ago

Evaluate lim x approaches to 2 : (sqrt(6-x)-2)/(sqrt(3-x)-1)

solong [7]

Answer:

The answer is "0.5".

Step-by-step explanation:

Given:

$\to \lim_{x\to 2} \frac{(\sqrt{(6-x)}-2)}{(\sqrt{(3-x)}-1)}\\\\ \to \lim_{x\to 2} \frac{\frac{d(\sqrt{(6-x)}-2)}{dx}}{\frac{(\sqrt{(3-x)}-1)}{dx}}\\\\$

$\to \lim_{x\to 2} \frac{\frac{-1}{2\sqrt{(6-x)} }}{\frac{-1}{2\sqrt{(3-x)}}}\\\\$

$\to \lim_{x\to 2} \frac{\sqrt{3-x}}{\sqrt{6-x}}\\\\ \to \frac{\sqrt{3-2}}{\sqrt{6-2}}\\\\ \to \frac{\sqrt{1}} {\sqrt{4}}\\\\ \to \frac{1}{2}\\\\ \to 0.5$