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Montano1993 [528]
3 years ago
15

Please help me answer this :) Explain your answer Will give brainiest

Mathematics
1 answer:
Greeley [361]3 years ago
8 0

Answer:

d. 770

Step-by-step explanation:

35 gallons per min

after 12 mins there is 350 gallons of water

35 x 12 = 420

how much water was in the tank BEFORE the valve was opened

350 + 420 = 770

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M 4= 111 what is m 5
Lina20 [59]
Hey,
If you meant that m^{4}= 111 than m^{5}=  \sqrt[4]{111^{5}} but you can't find the exact number.

Hope this helps :) and don't forget to choose the brainliest please :)
6 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} = <u>0.0083</u>.

3 0
3 years ago
Read 2 more answers
Describe how you would find the sum of 24,36and 13
Oksi-84 [34.3K]
I would add them together to get 73

Step-wise I would add the ones column together to get 13, carry the one over then add the tens up to get 70 - your answer is 73
3 0
3 years ago
Read 2 more answers
How to do this I don’t understand this please help
VashaNatasha [74]
I'm guessing you'd combine them. And the end equation would be

81x6y
7 0
3 years ago
An electric bulb is sold in a box measuring 5 cm by 4 cm by 4 cm. If the shopkeeper receives them in a carton measuring 50 cm by
katen-ka-za [31]

Answer:

250

Step-by-step explanation:

Bulb box and carton both are of cuboidal shape.

<u>For bulb </u><u>box</u><u> </u><u>the </u><u>d</u><u>imensions</u><u> </u><u>are:</u>

  • l = 5 cm, w = 4 cm, h = 4 cm

  • V_{Bulb\: box} =lwh

  • \implies V_{Bulb\: box} =5(4)(4)

  • \implies V_{Bulb\: box} =80\: cm^3

<u>For </u><u>Carton </u><u>the dimensions are</u><u>:</u>

  • l = 50 cm, w = 20 cm, h = 20 cm

  • V_{Carton} =lwh

  • \implies V_{Carton} =50(20)(20)

  • \implies V_{Bulb\: box} =20,000\: cm^3

To find the number of bulbs packed in the carton, divide the V_{Carton} by V_{Bulb\: box}

  • Number \:of \:bulbs =\frac{V_{Carton}}{V_{Bulb\: box}}

  • \implies Number \:of \:bulbs =\frac{20,000}{80}

  • \implies Number \:of \:bulbs = 250

So, 250 bulbs will be packed in one carton.

3 0
2 years ago
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