Someone listed all the numbers from 0 to in increasing order. then he crossed out all the numbers where some other digit(s) besides 0 and 1 were used.

how many are left?

32

Step-by-step explanation:

We need digits with only 0's and 1's.

There are 5 forms which we have to consider.

We have to consider 1-, 2-, 3-, 4-, 5-digit numbers.

1-digit number

For 1 digit number there is only 2 possibilities (0,1).

2-digit number

For 2 digit number the first digit will be fixed i-e 1 because if we take 0 to be the first digit it will reduce to one digit number, so first digit has to be 1. Now in the second digit place we have two possibilities 0 and 1.

So the total possible combination for 2-digit number is 1*2=2.

3-digit number

For 3 digit number first place is fixed with second and third having two possibilities giving us the combination for 3-digit number as 1*2*2=4.

4-digit number

Similarly for four digit number first place is fixed with the rest three with the possibility of two. So possible combination is 1*2*2*2=8.

5-digit number

For 5 digit we have the combination as 1*2*2*2*2=16.

So the total numbers left are 2+2+4+8+16=32.

the probability is 0

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°