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the probability that
the number of three-digit numbers with distinct digits that can be formed using the digits 1, 2,3,5,8, and 9 is
both the first digit and the last digit of the three-digit number are even numbers is
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Step-by-step explanation:

There are 6 digits given: 1,2,3,5,8 and 9

But we are only taking three at a time.

= 6!/3!

= 6*5*4*3*2/3*2

= 6*5*4

= 120

The first and last digits can only be even if the number takes the one of the forms "2 _ 8" or "8 _ 2". The middle number can be any of the remaining four.

This means the probability of getting a number beginning and ending with an even digits is:

= 8/120

= 1/15

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

i think it is 18

step-by-step explanation: