Please help asap 25 pts

In an examination, 80 students secured

faust18 [17]

Number of students who secured first class in English only is 20

Step-by-step explanation:

Step 1: Find the number of students who secured first class in English only.

Total number of students who secured first class in English or Maths, T = 80

Number of students who secured first class in Maths, M = 50

Number of students who secured first class in both English and Maths, EM = 10

⇒ Let the number of students who secured first class in English only be x.

Total students with first class, T = M + EM + x

⇒ 80 = 50 + 10 + x

⇒ 80 = 60 + x

∴ x = 80 - 60 = 20

3 0

2 years ago

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Debora [2.8K]

Answer:

6

Step-by-step explanation:

This problem centers on recognizing the relationship between fractions and repeating decimals.

Note the following:

$\frac{1}{3}=0.33333333333333333333333333333333333333333333333...$

or more simply $\frac{1}{3}=0.\bar{3}$

This means that $2\frac{1}{3}=2.\bar{3}$

Going back to the original expression, $3\frac{2}{3}+2.\bar{3}$ simplifies to $3\frac{2}{3}+2\frac{1}{3}$

Next, adding mixed numbers. For adding mixed numbers, whole number parts and fractional parts can be added separately (this is because of the associative and commutative properties of addition)... because

$3\frac{2}{3}+2\frac{1}{3}$ means the same thing as

$(3+\frac{2}{3})+(2+\frac{1}{3})$

Changing the order, we can write it as

$3+2+\frac{2}{3}+\frac{1}{3}$

Adding the whole number parts, 3+2 is 5.

Next, adding the fraction parts:

For adding fractions, one must have a common denominator (the bottom part of the fraction must be the same). Once they are the same, the denominator just sort of "hangs out" until the end without changing, and the numerators (the tops) are added together.

Fortunately, the denominators are already the same. So, add the numerators, and the denominator will stay as a 3.

$\frac{2}{3}+\frac{1}{3}$