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3 years ago

Joe ate twice as many pieces of Halloween candy as Ed. Together they ate 48 pieces. How many pieces of candy did ed eat ?

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

4 0

Answer:

Joe 2 : 1 Ed

2+1 = 3

48/3 = 16

Joe 32 : 16 Ed

Ed ate 16 pieces

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I need help<br> Geometry

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3 years ago

If you roll two number cubes, what is the probability of their sum being 9 or higher?

aalyn [17]

Answer:

$Pe = \frac{10}{36}$

Step-by-step explanation:

Well, a standard cube number (dice) has numbers from 1 to 6. So the combinations that give 9 or higher, are 3-6, 6-3, 4-5, 5-4, 4-6, 6-4,5-6, 6-5, 5-5, 6-6. We will call this favorable cases (Fc) = 10

Yet, the total number of cases (T) is 6 different number by 6 different combinations (1-1 , 1-2 , 1-3 ..... 6-6), 6 x 6= 36 = T.

So, in order to calculate the probability of an event (10/36), you can use the following formula:

$Pe = \frac{Fc}{T}$

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3 years ago

4(x+2)>60 I need help

IrinaVladis [17]

The answer is x>13. First you must simplify both sides of the inequality. Which is now 4x +8>60. Then you must subtract 8 from both sides and finally divide both sides by 4.
4x + 8 -8> 60 - 8
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4 divided by 52 is 13. The final answer is x>13.

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3 years ago

Read 2 more answers

An auditorium has rows of seats with 8 seats in each row. Kayla knows there are at least 70 seats but fewer than 340 seats in th

Vera_Pavlovna [14]

Answer:

The possible number of rows can be 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, or 42.

Step-by-step explanation:

First of all, if there are 70 seats, there should be 9 rows. (9*8 =72, which has 70 seats). And, 340 seats are there, so 340/8 = 42.5, i.e. 42 rows can be there at maximum. So, the actual number of rows can be between 9 and 42. Therefore the possible number of rows can be 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, or 42.

Hope this helps. :)

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3 years ago

The given lengths are two sides of a right triangle. All three side lengths of the triangle are integers and together form a Pyt

Ludmilka [50]

12 & 35 Third side is hypotenuse = 37
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20 & 29 Third side is a leg = 21

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3 years ago