What is 7 and 1/4% of 89.75?

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

3 0

6.506875 is the answer because 7.25% out of 100 is .0725 and 89.75 times that is 6.506875

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Two pentagons are similar. the area of one of the Pentagon is 9 times that the other. determine the ratio of the length of the c

kolbaska11 [484]

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2 years ago

A student bought 3 boxes of pencils and 2 boxes of pens for $6. He then bought 2 boxes of pencils and 4 boxes of pens for $8. Fi

Annette [7]

Let c and n represent the numbers of pencil and pen boxes, respectively
.. 3c +2n = 6.00
.. 2c +4n = 8.00

You can halve the second equation and subtract it from the first to get
.. (3c +2n) -(c +2n) = 6.00 -4.00
.. 2c = 2.00
.. c = 1.00

Then
.. 1.00 +2n = 4.00 . . . . . half the original second equation with c filled n
.. 2n = 3.00
.. n = 1.50

A box of pencils costs $1.00; a box of pens costs $1.50.

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3 years ago

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Kate and Melisa both went to the same store to buy eggs and bread for their families. Kate bought 3 cartons of eggs and 5 loaves

VMariaS [17]

Write an equation system based on the problem
For an instance, e stands for the price of a carton of eggs, and b stands for the price of a loaf of bread

"Kate bought 3 cartons of eggs and 5 loaves of bread for $30.07" could be written as 3e + 5b = 30.07 (first equation)
"Melissa bought 2 cartons of eggs and 2 loaves of bread for $14.74" could be written as 2e + 2b = 14.74 (second equation)

Solve the equation system using elimination method
Eliminate b to find e. To eliminate b, we should equalize the coefficient of b.
3e + 5b = 30.07 (multiplied by 2)
2e + 2b = 14.74 (multiplied by 5)
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3 years ago

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Flura [38]

Answer:

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Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

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