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2 years ago

You are going to draw 4 cards from a standard deck of 52 cards. How many different combinations of cards could you draw?

Mathematics

2 answers:

zloy xaker [14]2 years ago

5 0

Answer:

Step-by-step explanation:

To start with we can consider how many possibilities there are for the first card which is obviously 52, as any one of the cards in the deck can be found on the top of the pile. The next card is a bit trickier as there are only 51 possible cards that it could be as the first card takes up a spot. The third card now only has 50 possible cards it could and this continues all the way down to the last card.

is that good

Firlakuza [10]2 years ago

5 0

You can make 208 different combinations

Explanation: To find the combinations you can multiply 52 and 4 and you should get 208

You might be interested in

Does -3(x-4)= -3x=12 have one solution, many solutions, or no solutions at all?

TEA [102]

Answer:

If the expression is:

-3x + 12 = -3x + 12, infinite solutions

If the expression is:

-3x + 12 = -3x - 12, no solutions

Step-by-step explanation:

Step 1: Distribute

-3x + 12 = 3x = 12

If both lines have the same slope and same y-int, they are the same line and therefore have infinite solutions.

If both lines have the same slope but different y-int, then they are parallel and therefore have no solutions.

4 0

2 years ago

To travel 80 miles, it takes Sue, riding a moped, 2 hours less time than it takes Doreen to travel 60 miles riding a bicycle. Su

andriy [413]

Answer:

Time taken by Doreen is 6 hours and speed is 10 miles per hour.

Time taken by Sue is 4 hours and speed is 20 miles per hour.

Step-by-step explanation:

Let the speed of Doreen be x

According to the question speed of Sue is = x+10

time taken By Sue to cover 80 miles = $\frac{80}{x+10}$

time taken by Doreen to travel 60 miles = $\frac{60}{x}$

According to question Sue take two hours less than Doreen takes

therefore

$\frac{60}{x}$ - $\frac{80}{x+10}$ =2

$\frac{60(x+10)-80x}{x(x+10)}$ =2

60(x+10) -80x = 2(x(x+10)

60x+600-80x = $2x^2+20x\\$

simplifying it ,we get

$2x^2+40x-600=0\\$

Dividing both sides by 2 ,we get

$x^2+20x-300=0\\$

solving it for x ,we get

(x+30)(x-10) =0

x =-30 which is not possible

x =10 miles per hour

Speed of Doreen = 10 miles per hour

Speed of Sue = 10+10 = 20 miles per hour

Time taken by Doreen = 60 divided by 10 = 6 hours

Time taken by Sue = 80 divided by 20 = 4 hours

5 0

3 years ago

Convert 7.87% to a decimal

juin [17]

Your answer is 0.0787

8 0

2 years ago

For what values of b are the vectors [−18, b, 9] and [b, b2, b] orthogonal? (Enter your answers as a comma-separated list. If an

bearhunter [10]

Answer:

Therefore the given vectors are orthogonal for b = 0,±3.

Step-by-step explanation:

If $\vec a$ and $\vec b$ are two vectors orthogonal, then the dot product of $\vec a$ and $\vec b$ will be zero.

i.e $\vec a. \vec b =0$

If $\vec a = x_1\hat i+y_1\hat j +z_1\hat k$ and $\vec b = x_2\hat i+y_2\hat j +z_2\hat k$

$\vec a. \vec b=( x_1\hat i+y_1\hat j +z_1\hat k).( x_2\hat i+y_2\hat j +z_2\hat k)$

$=x_1 x_2+y_1y_2+z_1z_2$

Given two vectors are (-18,b,9) and (b,b²,b)

Let

$\vec P= -18 \hat i+b\hat j +9 \hat k$

and

$\vec Q = b \hat i+b^2 \hat j +b\hat k$

Therefore,

$\vec P.\vec Q$

$=( -18 \hat i+b\hat j +9 \hat k).( b \hat i+b^2 \hat j +b\hat k)$

=(-18).b+b.b²+9.b

= -18b+b³+9b

= b³-9b

Since $\vec P$ and $\vec Q$ are orthogonal. Then $\vec P.\vec Q$ = 0.

Therefore,

b³-9b= 0

⇒b(b²-9)=0

⇒b =0 or b²=9

⇒b=0 or b =±3

Therefore the given vectors are orthogonal for b = 0,±3.

8 0

3 years ago

¿De cuántas formas podemos cambiar un billete de 5000 por monedas de 500 y de 200,si tenemos q utilizar al menos de cada denomin

NARA [144]

Answer:

Existen 6 formas posibles de cambiar un billete de 5000 por monedas de 200 y 500.

Step-by-step explanation:

En primer lugar, se determina el mínimo común múltiplo de 500 y 200. Si se multiplica a 500 por 2 y a 200 por 5, se encuentra que el mínimo común múltiplo es 1000. Entonces, se encuentra las siguientes dos conclusiones:

1) Se requiere dos monedas de 500 para obtener 1000.

2) Se requiere cinco monedas de 200 para obtener 1000.

Para determinar todas las formas posibles sin considerar el orden de las monedas, se requiere estudiar tres escenarios:

(i) Todas las monedas son de 200.

(ii) Todas las monedas son de 500.

(iii) Hay monedas de 200 y 500.

Caso 1 - Todas las monedas son de 200:

Es evidente que solo existe una forma.

Caso 2 - Todas las monedas son de 500:

Es evidente que solo existe una forma.

Caso 3 - Hay monedas de 200 y 500:

Existen las siguientes formas:

- 1000 en monedas de 200, 4000 en monedas de 500.

- 2000 en monedas de 200, 3000 en monedas de 500.

- 3000 en monedas de 200, 2000 en monedas de 500.

- 4000 en monedas de 200, 1000 en monedas de 500.

En resumen, existen 4 escenarios.

Finalmente, existen 6 formas posibles de cambiar un billete de 5000 por monedas de 200 y 500.

4 0

3 years ago