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2 years ago

You roll a six-sided die twice. What is the probability of rolling a 2 and then an odd number?

Mathematics

2 answers:

shusha [124]2 years ago

5 0

Answer:

1/12

Step-by-step explanation:

possible outcome for a die = 6

Probability of rolling a 2 P(2) = 1/6

probability of rolling a odd number = no of odd numbers/ total possible outcome

P(odd) = 3/6 =1/2

probability of rolling a 2 and then an odd number = 1/6 x 1/2 = 1/12

Oduvanchick [21]2 years ago

3 0

Answer:

1/3

Step-by-step explanation:

What is the probability of rolling a 2 and then an odd number?

1, 2, 3, 5

4/12 bc u are rolling it twice

4/12 divided by 4 = 1/3

hope im correct and hope i helped

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Solve the following function for y when x=5 round to the nearest tenth

Wittaler [7]

In the given question , we have a function given , which is

$y= 13(0.5^x)$

And we need to find the value of y when x=5.

So we substitute 5 for x and solve for y, that is

$y=13(0.5^5) = 0.40625$

And on rounding to the nearest tenths, we will get

$y=0.4$

So when x=5, y=0.4

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2 years ago

How many tens are in the number 1,384

Andre45 [30]

There are 8 tens in the number 1,384.

Each digit in the number has an equivalent value based on its placement in the number.

1,384 where:

1 is in the thousands value
3 is in the hundreds value
8 is in the tens value
4 is in the ones value

The extended form of 1,384 is:

1 x 1000 + 3 x 100 + 8 x 10 + 4 x 1 = 1,384

1,000 + 300 + 80 + 4 = 1,384

8 x 10 = 80 ; shows that there are 8 10s in the number with a product of 80.

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3 years ago

Read 2 more answers

The coordinates of point A are (-6, -10) and the coordinates of point B are (x,-4). The distance between A to point B is 10 unit

vlabodo [156]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-10})\qquad B(\stackrel{x_2}{x}~,~\stackrel{y_2}{-4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ 10=\sqrt{[x-(-6)]^2+[-4-(-10)]^2}\implies 10=\sqrt{(x+6)^2+(-4+10)^2} \\\\\\ 10^2=(x+6)^2+(6)^2\implies 100=x^2+12x+36+36 \\\\\\ 100=x^2+12x+72\implies 0=x^2+12x-28 \\\\\\ 0=(x+14)(x-2)\implies x= \begin{cases} -14\\ \boxed{2} \end{cases}$

because B is on the IV Quadrant, the x-coordinate must be positive.

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2 years ago

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

$(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0$

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

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2 years ago

If the length of two sides of a triangle are 7 and 10 the length of the third side may be

Sonja [21]

Answer:

D. 4

Step-by-step explanation:

Let's look at all the options first. All of the options are below 7 and 10, hence we can deduce that it is not the hypotenuse. Looking at 7 and 10, we see that 10 is the larger number, and hence we know that 10 is the hypotenuse.

Hence, as the other two sides added up in a triangle must be larger than the hypotenuse, the only available option is 4, since 4 + 7 = 11 > 10.

Hope this helped!

6 0

2 years ago

You roll a six-sided die twice. What is the probability of rolling a 2 and then an odd number?​

You roll a six-sided die twice. What is the probability of rolling a 2 and then an odd number?