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3 years ago

Help me please! This is urgent.

Mathematics

2 answers:

DochEvi [55]3 years ago

7 0

Answer: W

Step-by-step explanation:

pychu [463]3 years ago

3 0

Answer:

I think W or Z but not really sure so

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Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

Read 2 more answers

The quotient of n and 6

Shalnov [3]

Answer:

The quotient of n and 6 is n/6, as quotient means divide.

for example, if it was "the quotient of 24 and 6", it would be 24/6, which is 4.

5 0

3 years ago

Find the next term is in the explicit and recursive rule for the in term of the sequence. NO LINKS!!!

BARSIC [14]

9514 1404 393

Answer:

5) 729, an=3^n, a[1]=3; a[n]=3·a[n-1]

6) 1792, an=7(4^(n-1)), a[1]=7; a[n]=4·a[n-1]

Step-by-step explanation:

The next term of a geometric sequence is the last term multiplied by the common ratio. (This is the basis of the recursive formula.)

The Explicit Rule is ...

$a_n=a_1\cdot r^{n-1}$

for first term a₁ and common ratio r.

The Recursive Rule is ...

a[1] = a₁

a[n] = r·a[n-1]

5. First term is a₁ = 3; common ratio is r = 9/3 = 3.

Next term: 243×3 = 729

Explicit rule: an = 3·3^(n-1) = 3^n

Recursive rule: a[1] = 3; a[n] = 3·a[n-1]

6. First term is a₁ = 7; common ratio is r = 28/7 = 4.

Next term: 448×4 = 1792

Explicit rule: an = 7·4^(n-1)

Recursive rule: a[1] = 7; a[n] = 4·a[n-1]

7 0

3 years ago

Write 34,900 in scientific notation

Rzqust [24]

Answer:It is

3.49

⋅

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Create a matrix that is equal to F+G. The first matrix below is named F and the second matrix below is named G. Name the new mat

likoan [24]

F+G:

$F+G=\begin{bmatrix}{-1.8} & {-8.6} & {} \\ {2.85} & {-1.4} & {} \\ {-1.8} & {5.1} & {}\end{bmatrix}+\begin{bmatrix}{1.32} & {-1.9} & {} \\ {2.25} & {0.0} & {} \\ {-6.2} & {1.4} & {}\end{bmatrix}$

Then, add the elements that occupy the same position:

$H=\begin{bmatrix}{-1.8+1.32} & {-8.6+(-1.9)} & {} \\ {2.85+2.25} & {-1.4+0.0} & {} \\ {-1.8+(-6.2)} & {5.1+1.4} & {}\end{bmatrix}$

Solve

$H=\begin{bmatrix}{-0.48} & {-10.5} & {} \\ {5.1} & {-1.4} & {} \\ {-8} & {6.5} & {}\end{bmatrix}$

So, we find the element at address h31:

$H=\begin{bmatrix}{h11} & {h12} & {} \\ {h21} & {h22} & {} \\ {h31} & {h32} & {}\end{bmatrix}$

In this case, position h31 is - 8.0

8 0

1 year ago