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Eduardwww [97]
3 years ago
9

A blue die and a red die are thrown. B is the event that the blue comes up with a 6. E is the event that both dice come up even.

Write the sizes of the sets |E ∩ B| and |B|a. |E ∩ B| = ___b. |B| = ____
Mathematics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

Size of |E n B| = 2

Size of |B| = 1

Step-by-step explanation:

<em>I'll assume both die are 6 sides</em>

Given

Blue die and Red Die

Required

Sizes of sets

- |E\ n\ B|

- |B|

The question stated the following;

B = Event that blue die comes up with 6

E = Event that both dice come even

So first; we'll list out the sample space of both events

B = \{6\}

E = \{2,4,6\}

Calculating the size of |E n B|

|E n B| = \{2,4,6\}\ n\ \{6\}

|E n B| = \{2,4,6\}

<em>The size = 3 because it contains 3 possible outcomes</em>

Calculating the size of |B|

B = \{6\}

<em>The size = 1 because it contains 1 possible outcome</em>

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+6

Step-by-step explanation:

Just do 0 -(-6).

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2 years ago
Yet another math question
bonufazy [111]

Answer:

69.5

Step-by-step explanation:

Formula = 1/2 ( b1 + b2)h

base 1 + base 2 = 220

7645 x 2 = 14000 + 1200 + 80 + 10 = 15290

15290 / 220 =  764.5

H = 69.5

Lets check

1/2(220 x 69.5)

1/2(15290) = 7645

correct

If my answer is incorrect, pls correct me!

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-Chetan K

6 0
2 years ago
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How to know if a function is periodic without graphing it ?
zhenek [66]
A function f(t) is periodic if there is some constant k such that f(t+k)=f(k) for all t in the domain of f(t). Then k is the "period" of f(t).

Example:

If f(x)=\sin x, then we have \sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x, and so \sin x is periodic with period 2\pi.

It gets a bit more complicated for a function like yours. We're looking for k such that

\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5

Expanding on the left, you have

\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2

and

1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5

It follows that the following must be satisfied:

\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}

The first two equations are satisfied whenever k\in\{0,\pm4,\pm8,\ldots\}, or more generally, when k=4n and n\in\mathbb Z (i.e. any multiple of 4).

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It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when k is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2

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