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3 years ago

When a die is rolled what is the probability of rolling a 2 or a 5?

Mathematics

2 answers:

Travka [436]3 years ago

4 0

$\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{2}{6}=\dfrac{1}{3}$

Elis [28]3 years ago

4 0

1/3 because there is a 1/6 chance of getting any number. So 1/6 + 1/6 = 2/6 which is simplified to 1/3

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Answer: 2
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3 years ago

What best describes the area of a polygon

never [62]

Answer:

The area of any regular polygon is given by the formula: Area = (a x p)/2, where a is the length of the apothem and p is the perimeter of the polygon. Plug the values of a and p in the formula and get the area. As an example, let's use a hexagon (6 sides) with a side (s) length of 10.

The area of a polygon is the two-dimensional set of all points surrounded by the sides of the polygon.

If you're looking for an equation, it varies based on the number of sides and the shape of the polygon.

Step-by-step explanation:

Apothem

A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). To find the area of a regular polygon, you use an apothem — a segment that joins the polygon’s center to the midpoint of any side and that is perpendicular to that side (segment HM in the following figure is an apothem).

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3 years ago

P(A) = .6, P(B) = .3, P(Β | A) = .5

Reika [66]

Answer:

Step-by-step explanation:

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2 years ago

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natta225 [31]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$

$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

The final value of the converges series for all x.

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3 years ago