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Harlamova29_29 [7]
3 years ago
15

Partial qoutients 231÷11

Mathematics
1 answer:
marta [7]3 years ago
8 0

Answer:

21

Step-by-step explanation:

We are to find partial quotients of 231 ÷ 11

In other words the question asks: How many 11s are there in 231

First, there are twenty 11s in 220 i.e 20 × 11 = 220

Then, subtract 220 from 231 to get 11

Finally ask yourself: How many 11s are in eleven? Definitely its only one 11.

So there are 20 + 1 elevens in 231

Or, there are 21 elevens in 231

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Out of the 400 students in a final year in secondary school, 300 are offering biology and 190 are offering chemistry. I) how man
iren2701 [21]

Answer:

I) how many student are offering both subject , if only 70 students are offering neither biology nor chemistry?

= 160 students

II) how many students are offering at least one biology and chemistry?

= 330 students

Step-by-step explanation:

I) how many student are offering both subject , if only 70 students are offering neither biology nor chemistry?

Total number of student n ( B ∪ C) = 400

Students offering biology n(B) = 300

Students offering chemistry n(C) = 190

Students not offering biology nor chemistry = 70

Students offering both biology and chemistry n ( B ∩ C) = ??

n ( B ∪ C) - Students neither offering biology nor chemistry = n(B ) + n ( C ) - n ( B ∩ C)

400 - 70 = 300 + 190 - n ( B ∩ C)

330 = 490 - n ( B ∩ C)

n ( B ∩ C) = 490 - 330

n ( B ∩ C) = 160

Therefore, students offering both Biology and Chemistry are 160 students.

II) how many students are offering at least one biology and chemistry?

n(B' ) + n ( C') + n ( B ∩ C)

n(B') = Students offering biology only

= n(B) - n ( B ∩ C)

= 300 - 160 = 140

n(C') = Students offering chemistry only

n(C) - n ( B ∩ C)

= 190 - 160

= 30

(300 - 160) + (190 - 160) + 160

140 + 30 + 160

= 330 students

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What is the solution to this system of equations? x+y=2 y-7=2 x+y-7=3
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