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MariettaO [177]
3 years ago
15

Rihanna

Mathematics
2 answers:
Daniel [21]3 years ago
7 0
The answer to your question is 71%
ANEK [815]3 years ago
3 0

First you would subtract 1.5 from 2.1,

2.1-1.5=0.6

0.6=60%

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masya89 [10]
The percent would be 20%
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Can someone help me with this, It's a bit confusing for me?
daser333 [38]

Step-by-step explanation:

use the Pythagoras theorem

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3 years ago
370 College freshmen were interviewed. 200 were registered for an Advanced Math class. 270 were registered for an English class
frozen [14]

Answer:

Step-by-step explanation:

To aid in understanding, draw three circles...

100 were registered for both Math and English.

50 were registered for both Sport and English but not Math.

70 were registered for Math, Sport, and English.

From here we will be able to calculate for those that signed up for only English class: Total number of students that signed up for english class

a. none signed up only for Maths (refer to attached document)

b. How many signed up for neither Math nor English meaning they signed up for only sports which is 70 from the venn diagram in the attachment.

c. How many signed up for an English class and at least one other class?

From the venn diagram:

we could have an English class and maths class - 100

we could also have an a English class and sports - 50

Thus total is 150.

d. What is the total fee for all the students enrolled in classes

From the diagram, 120 (english only and sport only) enrolled for only one class = 120 x $100 = $12,000

From the diagram, 180 (english and sport, maths and sport and english and maths) enrolled for two classes = 180 x $150 = $27,000

From the diagram, only 70 enrolled for the three classes = 70 x $200 = $14,000

Total fees = $12,000 + $27,000 + $14,000 = $53,000

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4 years ago
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Romashka-Z-Leto [24]

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A way to do it by hand is

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8 * 5

7 * 6

stop here because 6 * 7 and 5 *8 are already listed

This gives you 6 products then multiply those 6 products by pairs which gives you three pairs which gives you three products then compute a pair then multiply by last product.

Easy math but a lot of written work

3 0
3 years ago
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