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2 years ago

A pair of standard dice is rolled. Find the probability that the sum of the two dice is greater than 1.

Mathematics

1 answer:

sleet_krkn [62]2 years ago

4 0

Answer:

1/1 chance

Step-by-step explanation:

The lowest sum the die could add up to is 2. 2>1

So that means that on every roll, the sum of the two die will add to a number greater than one.

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What is the difference between the greatest and least amounts of rainfall

drek231 [11]

Answer:

- The greatest amount of rainfall is represented by the greatest number

- The least amount of rainfall is represented by the least number

Step-by-step explanation:

5 0

3 years ago

How do you subtract nega

sergey [27]

Well think of it like this. lets say your teacher made you do this problem -48 - 10. I would just act like 48 is not an negative and add 10. After that I would put the negative back. So the answer would be -58. :P

3 0

4 years ago

Read 2 more answers

Tolong bantu jawab please, makasih sebelumnya!

sweet [91]

Answer:

2x² - 3x - 2

Explain:

Use the FOIL method: (a +b)(c +d) = ac + ad +bc +bd.

2x² + x - 4x - 2

Collect like terms.

2x² + (x - 4x) - 2

Simplify.

2x² - 3x - 2

6 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

Which function is negative for the interval (–1, 1]?

Taya2010 [7]

I think it’s the second one

7 0

3 years ago

Read 2 more answers