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vitfil [10]
3 years ago
13

At marco's school,5/8 of the students are in the band.What percent of the student are in the band?

Mathematics
2 answers:
sammy [17]3 years ago
6 0
62.5 percent of students are in the band because you cross multiply and then divide 500 by 8 giving you 62.5 over 100
Neko [114]3 years ago
4 0

Answer:

62.5%

Step-by-step explanation:

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Write a number sentence to represent the sum modeled on the number line. In two or more complete sentences, explain your reasoni
Afina-wow [57]

Answer:

0 + (-2.5) + 5

Step-by-step explanation:

from 0 to -2.5 then to 5 so add 0 + (-2.5) + 5

4 0
3 years ago
Decimals between 10 and 15 with an interval of .75 between each pair of decimals
d1i1m1o1n [39]
This is a sum that needs understanding the starting and the ending number first. then you should understand that between the starting and the ending number there has to be pairs of numbers with a difference of 0.75.
The numbers are:
10.75, 11.50, 12.25, 13.00, 13.75, 14.50.
6 0
4 years ago
How to write 3482000000 in scientific notation
Ber [7]
To write 3482000000 in scientific notation. Put down the first number (3) and all numbers that come after it (not 0) put after a decimal point. Then count all of the numbers after the decimal points including 0's and that's your exponent. Like so,
3.482*10^9

Hope this helps!=)
3 0
3 years ago
Read 2 more answers
PLEASE HELP ME 3 friends are camping in the woods, Bert, Ernie and Elmo. They each have their own
fredd [130]

The answer is B

Reason:

Ernie ?

Bert 30°

Elmo 105°

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6 0
3 years ago
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One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r
Flura [38]

Answer:

12 possibilities

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.

3 0
3 years ago
Read 2 more answers
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