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3 years ago

the ratio of boys to girls in a class is 4 : 5 how many girls are there in a class if 12 boys are in the class

Mathematics

2 answers:

bija089 [108]3 years ago

5 0

Hey! Pretty stranger :D

AsnwEr:

Number of girls in class = 15

GivEn:

The ratio of no. of boys to no. of girls in a class is 4:5.
Number of boys in class = 12

To find:

Number of girls in class?

Solution:

☯ Let number of girls be x.

Given that,

The ratio of no. of boys to no. of girls in a class is 4:5.

Therefore,

→ 12/x = 4/5

→ x = 12 × 5/4

→ x = 60/4

→ x = 15

∴ Hence, number of girls in a class is 15.

Brums [2.3K]3 years ago

3 0

Answer:

Step-by-step explanation:

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The image of the floor dimensions is missing, so i have attached it.

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3 years ago

Prove that u(n) is a group under the operation of multiplication modulo n.

katrin2010 [14]

Answer:

The answer is the proof so it is long.

The question doesn't define u(n), but it's not hard to guess.

Group G with operation ∘

For all a and b and c in G:

1) identity: e ∈ G, e∘a = a∘e = a,

2) inverse: a' ∈ G, a∘a' = a'∘a = e,

3) closed: a∘b ∈ G,

4) associative: (a∘b)∘c = a∘(b∘c),

5) (optional) commutative: a∘b = b∘a.

Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.

If n isn't prime, we exclude from the group all integers which share factors with n.

Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).

Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).

Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.

Associative and Commutative:

(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)

a∘b = b∘a because ab = ba.

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