Nashan randomly chooses 6 positive integers a, b, c, d, e, f. Find the probability that 2 ^a + 2^b + 2^c + 2^d + 2^e + 2^f is divisible by 5.

0.20

Step-by-step explanation:

we have that:

a number is only divisible by 5 if:

the unit digit of the number is 0 or 5.

As we have:

2^a + 2^b + 2^c + 2^d + 2^e + 2^f = X

we can never have 5 as the last digit, so we only can have a zero as the last digit.

Now, we know that

2^a + 2^b + 2^c + 2^d + 2^e + 2^f can only end with even numbers, if we assume the same probability for all of them, we have 0, 2, 4, 6, 8

5 even numbers, and the sum will be divisible by 5 only if the last number is a zero, then the probability having a zero in the last number is equal to the number of times that zero appears in that set, divided the total number of elements in that set.

p = 1/5 = 0.20

the answer is 48m^2!

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b