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3 years ago

What is the probability of a 6 sided dice

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Answer:

probability = 1 ÷ 6 = 0.167, or 16.7 percent chance.

Step-by-step explanation:

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What will be the total cost of Devon's first year?

ELEN [110]

Answer: Q1: $8,050

Q2: $5,400

Q3: $2,650

Step-by-step explanation:

4 0

3 years ago

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Cuantos numeros enteros hay entre -3 y +3?

romanna [79]

Hay 6 números entre -3 y 3

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3 years ago

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If a stone is thrown down at 170 feet per second from a height of 1,020 feet, its height after t seconds is given by

andreyandreev [35.5K]

Step-by-step explanation:

It is given that,

Initial speed of the stone, u = 170 ft/s

It is thrown down from a height of 1020 feet.

We need to its height after t seconds. It can be calculated using second equation of kinematics as follows :

$h=ut-\dfrac{1}{2}gt^2+h_0$

h₀ is initial height, u = -170 ft/s as the stone is throwing down and g = -32ft/s²

Putting values, we get :

$h=-170t-\dfrac{1}{2}\times 32t^2+1020\\\\h=-16t^2-170t+1020$

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3 years ago

The after school craft center has 15 boxes of 64 crayons each. In 12 of the boxes , 28 of the crayons have not been used. All th

Korolek [52]

The <em><u>correct answer</u></em> is:

624 crayons.

Explanation:

In 12 of the boxes, 28 crayons have not been used; this leaves 64-28=36 crayons that have been used. 12(36) = 432 crayons have been used in these boxes.

3 full boxes have been used; this is 3(64) = 192 crayons.

Together this makes 432+192 = 624 crayons that have been used.

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3 years ago

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The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

3 years ago