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3 years ago

When you roll a standard number cube once, what is the probability of rolling a number divisible by 3?

Mathematics

1 answer:

sashaice [31]3 years ago

3 0

Answer:

$\frac{1}{3}$

Step-by-step explanation:

On a standard cube there are the numbers 1, 2, 3, 4, 5, 6

Only 2 are divisible by 3, that is 3 and 6, hence

P(divisible by 3 ) = $\frac{2}{6}$ = $\frac{1}{3}$

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The Kitchen committee purchased 76 boxes of cookies for Vacation Bible School,

iren2701 [21]

269 I think :) not sure

5 0

3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

The equation of a circle is given below.

BARSIC [14]

Given:

The equation of the circle is $x^2+(y+4)^2=64$

We need to determine the center and radius of the circle.

Center:

The general form of the equation of the circle is $(x-h)^2+(y-k)^2=r^2$

where (h,k) is the center of the circle and r is the radius.

Let us compare the general form of the equation of the circle with the given equation $x^2+(y+4)^2=64$ to determine the center.

The given equation can be written as,

$(x-0)^2+(y+4)^2=64$

Comparing the two equations, we get;

(h,k) = (0,-4)

Therefore, the center of the circle is (0,-4)

Radius:

Let us compare the general form of the equation of the circle with the given equation $x^2+(y+4)^2=64$ to determine the radius.

Hence, the given equation can be written as,

$x^2+(y+4)^2=8^2$

Comparing the two equation, we get;

$r^2=8^2$

$r=8$

Thus, the radius of the circle is 8

3 0

4 years ago

what is the equation subtracting 7 from the product of 4 and a number results in the number added to 29

vitfil [10]

That would be:-

4x - 7 = x + 29

the solution:-

3x = 29 + 7

3x = 36

x = 12

7 0

3 years ago

What's the square root of 15 divided by 5 square root of 20?

Arturiano [62]

Answer:

√3/10 = 0.1732

Step-by-step explanation:

√15 / 5√20 = (√5 x √3) / 5(√5 x √4) = √3 / (5 x√4) = √3 / 10

7 0

3 years ago