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melamori03 [73]
3 years ago
10

2. What is the product of 4 2/3 and 11 1/4?

Mathematics
1 answer:
Lerok [7]3 years ago
3 0
4 \frac{3}{4} ~*~ 11\frac{1}{4} =  \frac{14}{3} ~*~ \frac{45}{4} =  \frac{630}{12}=52 \frac{6}{12}  =[52 \frac{1}{2} ]
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Find volume of sphere in cubic yards whose radius is 5 yard<br><br> Use pi = 3.14​
enyata [817]

Answer:

523.333 cubic yards

Step-by-step explanation:

The formula for the volume of a sphere is  \frac{4}{3}\pi r^{3} where r is the radius.

The radius of the given sphere is r=5 yards and \pi =3.14.

Therefore, we can substitute in the radius and solve:

V_{sphere}= \frac{4}{3}\*\times3.14\times  5^{3} =523.333 cubic yards

4 0
2 years ago
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Answer:

yes

Step-by-step explanation:

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3 years ago
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Luba_88 [7]

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4 0
2 years ago
Middle School: Math (10 Points)
sasho [114]
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:

A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?

You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked

2) a - 2%  as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.  

b - 1,000,000/2500 = 400
400 packages are defective each year
5 0
3 years ago
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