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The cafeteria at a summer camp gives each camper 2/3 cup of juice for breakfast. This morning, 50 campers had juice for breakfas

qaws [65]

2/3 x 50/1 = 100/3 or 33 1/3

6 0

3 years ago

-3(2r-4) > 2(5-3r) Whats the inequality?

VLD [36.1K]

-3 (2r - 4) > 2 (5 - 3r)
-6r + 12 > 10 - 6r
12 > 10

There is no solution

4 0

3 years ago

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An indoor water park has two giant buckets that slowly fill with 1000 gallons of water before dumping it on the people below. On

Art [367]

Answer:

The next two times both buckets dump water at the same time are 4:54 P.M and 7:18 P.M.

Step-by-step explanation:

From the question, one bucket dumps water every 18 minutes, say bucket A;

and the other bucket dumps water every 16 minutes, say bucket B.

Also, It is currently 2:45 P.M. and both buckets dumped water 15 minutes ago,

that is, both buckets dumped water at 2:30 P.M.

For bucket A, the next times it will dump water will be:

2:48 P.M, 3:06 P.M, 3:24 P.M, 3:42 P.M, 4:00 P.M, 4:18 P.M, 4:36 P.M, 4:54 P.M, 5:12 P.M, 5:30 P.M, 5:48 P.M, 6:06 P.M, 6:24 P.M, 6:42 P.M, 7:00 P.M, 7:18 P.M, 7:36 P.M, 7:54 P.M, 8:12 P.M, 8:30 P.M etc.

For bucket B, the next times it will dump water will be:

2:46 P.M, 3:02 P.M, 3:18 P.M, 3:34 P.M, 3:50 P.M, 4:06 P.M, 4:22 P.M, 4:38 P.M, 4:54 P.M, 5:10 P.M, 5:26 P.M, 5:42 P.M, 5:58 P.M, 6:14 P.M, 6:30 P.M, 6:46 P.M, 7:02 P.M, 7:18 P.M, 7:34 P.M, 7:50 P.M etc.

From above, we observe that the next two times both buckets dump water at the same time are 4:54 P.M and 7:18 P.M.

Hence, the next two times both buckets dump water at the same time are 4:54 P.M and 7:18 P.M.

6 0

3 years ago

Really need help Hurryyyyyyy

hram777 [196]

Answer:

The Answer is B

Ive hade this test before

5 0

3 years ago

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 = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = 0.0083.

3 0

3 years ago

Read 2 more answers