How do you do this and what are the answers

Mathematics

1 answer:

scoray [572]3 years ago

5 0

I can answer 1 to 4...
The area of a circle can be found by $\pi r^{2}$ . Reminder: $\pi = 3.14$ (usually used when finding out the area of a circle)

1. $4^{2} = 16$
$\pi *16$ or $3.14 * 16 = 50.24$
$50.24/4=12.56$

2. $6^{2} = 36$
$\pi *36$ or $3.14 * 36 = 113.04$
$113.04/10=11.304$

(Not really sure if this is correct) 3. $2^{2}=4$
$\pi *4$ or $3.14 * 4$
$12.56/2.66666667=11.70999999$

4. $6^{2}=36$
$\pi *36= 113.04$
$113.04/6=18.84$

You might be interested in

Kim bought a hat. in California. The hat cost $15 before the 8% sales tax was applied. How much did Kim pay for the hat after ta

gizmo_the_mogwai [7]

Answer:

$16.2

Step-by-step explanation:

15 x .08 [wonder how i got .08? 8% = 0.8]

= 1.2

here's where most people mess up, they usually forget to add the tax to the original cost.

15 + 1.2=

$16.2

7 0

3 years ago

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).

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