Answer quickly pls

b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum

Mariana [72]

Answer:

${52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in ${52 \choose 7}$ ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in ${7 \choose 2}$ ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in ${45 \choose 7}$ . Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in ${7 \choose 2}$ ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in ${38 \choose 7}$ . And again, we have to choose which ones are the hidden ones, which can be chosen in ${7 \choose 2}$ ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in ${31 \choose 7}$ ways. And choosing which ones are the hidden ones for this player can be done in ${7 \choose 2}$ ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

${52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

7 0

3 years ago

Alan hikes 4.5 miles in 3 hours. Determine whether quantities (distance and time) vary directly or inversely and find the consta

Mrac [35]

V = x/t = 4.5 mil / 3 h = 1.5 mph
Since, v = 1.5 mph (or we can say that v is constant), distance and time vary directly and the constant of variation is 1.5 mph.

3 0

3 years ago

Can someone please help

ruslelena [56]

You can find the answer by substituting the x-values provided in the question and seeing if you get what the question says.

a satisfies all of the parameters.
b doesn’t satisfy g(-1), so it’s not b.
c doesn’t satisfy g(-1), so it’s not c.
d doesn’t satisfy g(-1), so it’s not d.

The correct answer is a.

6 0

3 years ago

April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3

vlada-n [284]

Answer:

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = $5\frac{1}{4}\ hrs$

$5\frac{1}{4}\ hrs$ can be Rewritten as $\frac{21}{4}\ hrs$

Number of Hours Carl worked on Math project = $\frac{21}{4}\ hrs$

Number of Hours Sonia worked on Math project = $6\frac{1}{2}\ hrs$

$6\frac{1}{2}\ hrs$ can be rewritten as $\frac{13}{2}\ hrs$

Number of Hours Sonia worked on Math project = $\frac{13}{2}\ hrs$

Number of Hours Tony worked on Math project = $5\frac{2}{3}\ hrs$

$5\frac{2}{3}\ hrs$ can be rewritten as $\frac{17}{3}\ hrs.$

Number of Hours Tony worked on Math project = $\frac{17}{3}\ hrs.$

Now Given:

April worked $1\frac{1}{2}$ times as long on her math project as did Carl.

$1\frac{1}{2}$ can be Rewritten as $\frac{3}{2}$

Number of Hours April worked on math project = $\frac{3}{2}$ $\times$ Number of Hours Carl worked on Math project

Number of Hours April worked on math project = $\frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs$

Also Given:

Debbie worked $1\frac{1}{4}$ times as long as Sonia.

$1\frac{1}{4}$ can be Rewritten as $\frac{5}{4}$ .

Number of Hours Debbie worked on math project = $\frac{5}{4}$ $\times$ Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = $\frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs$

Also Given:

Richard worked $1\frac{3}{8}$ times as long as tony.

$1\frac{3}{8}$ can be Rewritten as $\frac{11}{8}$

Number of Hours Richard worked on math project = $\frac{11}{8}$ $\times$ Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = $\frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs$

Hence We will match each student with number of hours she worked.

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

5 0

3 years ago

Read 2 more answers

12. The Opera is collecting monies in advance for ticket sales. The tickets cost $17.50 each.

Debora [2.8K]

Answer:

The number of tickets they have sold in advance are summed up with the following equation and solution:

Equation: $\frac{6842.50}{17.50}$ [the final answer should be in dollars]

Solution: The Opera sold 391 tickets in advance.

Step-by-step explanation:

First: Review some background information:

Let's see what the problem gives us to answer it. I'll bold out the important parts:

The Opera is collecting monies in advance for ticket sales. The tickets cost $17.50 each. If they have collected $6842.50 already, then how many tickets have they sold in advance?

We know how much the tickets cost and how much money they have earned from selling tickets, we can find out how much tickets they have sold.

Second: Solve the problem:

We can solve this problem using many ways, but I'm just going to pick one. We will use proportions to answer this. Now thoughts might be going through your head like this: What is this guy saying? What in the world is proportions? Well hopefully you learn from this example (Please don't take offense if you <em>do</em> know how to do proportions).

First we set up two fractions. One ticket costs $17.50, so we will use it to set up our first fraction: $\frac{1}{17.50}$ . We don't know how many tickets all together cost $6842.50 so we will use the variable: x. Now, we can set up our second fraction. x tickets costs $6842.50 so our second fraction will be: $\frac{x}{6842.50}$ . Now we must use cross multiplication. We multiply the numerator of the first fraction times the denominator of the second fraction, and then we multiply the denominator o the first fraction times the numerator of the second fraction. That's what we would do usually. But now, since we have a variable, we must use cross multiplication to solve for this variable. So we multiply the the numerator of the first fraction times the denominator of the other fraction and then we divide that answer times the denominator of the first fraction (You don't have to set it up like this).So let's solve this question:

$\frac{1}{17.50}$ $\frac{x}{6842.50}$

1 x 17.50 = 17.50

6842.50 ÷ 17.50 = 391

So our equation is 6842.50 ÷ 17.50, and our solution is 391 tickets

Topic: Proportions

6 0

3 years ago