Howard chose a candy from a bowl with 5 chocolate candies, 4 gummy candies and 6 hard candies. What is Howard's dependent probability of choosing a chocolate candy followed by a gummy candy?

Round to the nearest tenth of a percent, if necessary.

8.9%

Step-by-step explanation:

Here, we are to calculate the probability of Howard choosing a chocolate candy followed by a gummy candy.

The probability of selecting a chocolate candy = number if chocolate candy/ total number of candy

Total number of candy = 5 + 4 + 6 = 15

Number of chocolate candy = 5

The probability of selecting a chocolate candy = 5/15 = 1/3

The probability of selecting a gummy candy = number of gummy candies/total number of candies

Number of gummy candy = 4

The probability of selecting a gummy candy = 4/15

The probability of selecting a chocolate candy before a gummy candy = 1/3 * 4/15 = 4/45 = 0.088888888889

Which is same as 8.89 percent which is 8.9% to the nearest tenth of a percent

(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