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Alexxx [7]
3 years ago
11

Consider a standard deck of 52 playing cards, a randomly selected card from the deck, and the following events: R = red, B = bla

ck, A = ace, N = nine, D = diamond, and C = club.
Are D and C mutually exclusive?
Mathematics
1 answer:
Alika [10]3 years ago
6 0

Answer:

Events D and C are mutually exclusive.

Step-by-step explanation:

Two events D and C are mutually exclusive if

Pr(D\cup C)=Pr(D)+Pr(C)

Find these probabilities. In a standard deck of 52 playing cards there are 13 diamond cards and 13 club cards.

The probabilty of chosing diamond is

Pr(D)=\dfrac{13}{52}=\dfrac{1}{4}

The probabilty of chosing club is

Pr(C)=\dfrac{13}{52}=\dfrac{1}{4}

The probabilty of chosing diamond or club is

Pr(D\cup C)=\dfrac{13+13}{52}=\dfrac{1}{2}

Since

\dfrac{1}{2}=\dfrac{1}{4}+\dfrac{1}{4},

events D and C are mutually exclusive.

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Step-by-step explanation:

espero que ayude

5 0
3 years ago
Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. b)
zlopas [31]

We can express the recurrence,

\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}

in matrix form as

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}

By substitution,

\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}

and continuing in this way we would find that

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}

Diagonalizing the coefficient matrix gives us

\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

which makes taking the (n-2)-th power trivial:

\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

So we have

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}

and in particular,

a_n=\dfrac{29\left(2(-5)^{n-1}+5\cdot2^{n-1}\right)-47\left(-(-5)^{n-1}+2^{n-1}\right)}7

a_n=\dfrac{105(-5)^{n-1}+98\cdot2^{n-1}}7

a_n=15(-5)^{n-1}+14\cdot2^{n-1}

\boxed{a_n=-3(-5)^n+7\cdot2^n}

6 0
3 years ago
Find the surface area of each figure. Round your answers to the nearest hundredth, if necessary.
Ira Lisetskai [31]

Answer:

Step-by-step explanation:

The given figure is an hexagonal pyramid. It has a hexagonal base and six triangular lateral faces. The formula for determining the surface area of the pyramid is expressed as

Surface area = 3ab + 3bs

Where

a represents the apotherm which is the distance from each side of the hexagon to the center.

b represents the base length

s represents the slant height of the hexagonal base.

From the information given,

a = 5.2 cm

b = 6 cm

s = 12.2 cm

Surface area = (3 × 5.2 × 6) + (3 × 6 × 12.2) = 93.6 + 219.6 = 313.2 cm²

8 0
4 years ago
X + 54 = 58 what does "x" equal??
MaRussiya [10]
Hey there!

54 + 4 = 58

so x = 4

Hope this helps
Have a great day (:
4 0
3 years ago
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