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soldier1979 [14.2K]
3 years ago
8

PLZZZ HELP ME! help me

Mathematics
1 answer:
alexgriva [62]3 years ago
5 0

Answer:

1. 1/3

2. 4/9

3. 5/8

Step-by-step explanation:

1. 3/6 - 1/6 = 2/6 = 1/3

2. 7/9 - 3/9 = 4/9

3. 7/8 - 2/8 = 5/8

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Please help me with this question
Vladimir [108]

Answer: 20

Step-by-step explanation:

numbers belonging to the figure on the left are 2.5 times less than on the right so to find x multiply 8 by 2.5 and that is 20

7 0
2 years ago
What is the volume of a cylindrical bucket that is 35 cm tall and has a base with a diameter of 21 cm? Use pi= 22/7 and round yo
Natali [406]
The volume of a cylinder is given by the multiplication of the area of the base by the height of the cylinder.  π×r²(area of the base) × h (height of the cylinder) ⇔
\frac{22}{7}×(\frac{21}{2} )^{2}cm²×35cm=12127.5cm³
(r is half of the diameter. Because we are doing the square of r, the cm also are squared.)
Rounding to the nearest tenth is to round the result so that it maintains a number after the point - the result is 12127.5cm³
6 0
3 years ago
Read 2 more answers
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
you have an initial amount of 55$ in a savings account and you deposit an equal amount into your account each week. After 6 week
sveticcg [70]
55+added=133
added=amount addeed per week times 6=6x

55+6x=133
minus 55 both sides
6x=78
divide both sides by 6
x=13

add 13 per week

y=13x+55
3 0
3 years ago
The first three terms of a sequence are given. Round to the nearest thousandth (if necessary).
baherus [9]
The nth term is 6n+4 and the 40th term is 244
7 0
3 years ago
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