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3 years ago

Given that a x 60 = b what is the value of 4b/a

Mathematics

2 answers:

Marysya12 [62]3 years ago

5 0

Answer:

The value of $\frac{4b}{a}$ is 240

Step-by-step explanation:

We were given the equation: a x 60 = b

To find the value of 4b/a, we are going to follow the following step;

First, divide both-side of the equation by a, thus our equation becomes:

a x 60/a = b/a

On the left-hand side of the equation, 'a' at the numerator will cancel-out 'a' at the denominator. Our equation becomes;

60 = b/a

Then, the next step is to multiply 4 to both-side of the equation, our equation becomes

4 x 60 = 4 x b/a

240 = $\frac{4b}{a}$

$\frac{4b}{a}$ =240

Therefore, the value of $\frac{4b}{a}$ is 240

Ainat [17]3 years ago

3 0

Answer:

240

Step-by-step explanation:

Given

60a = b ( divide both sides by a )

60 = $\frac{b}{a}$ ( multiply both sides by 4 )

240 = $\frac{4b}{a}$

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Find the probability of the following events , when a dice is thrown once:

Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

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