The weights of bags filled by a machine are normally distributed with a standard deviation of 0.055 kilograms and a mean that can be set by the operator. At what level should the mean weight be set if it required that only 1% of the bags weigh less than 10.5 kilograms? Round the answer to 2 decimal places.

Step-by-step explanation:

Given that the  weights of bags filled by a machine are normally distributed with a standard deviation of 0.055 kilograms and a mean that can be set by the operator.

Let the mean be M.

Only 1% of the bags weigh less than 10.5 kilograms

i.e. P(X<10.5) = 0.01

corresponding Z value for P(Z<z) = 0.01 is -0.025

i.e. 10.5 = M-0.025(0.055)

Solve for M from the above equation

M =

Rounding off we get

10.50 kgs

Mean weight should be fixed as 10.50 kg.

15%

step-by-step explanation:

use the table and count the number of guesses that has two correct answers.

the following are two correct answers

10010, 01100, 11000

a total of 3 numbers having 2 correct guesses out of 20 questions.

probability of getting 2 correct guesses = the number of questions getting 2 correct answers / total number of questions.

= 3/20

to convert it into percent multiply by 100, we get

= 3/20 * 100

= 15%

the answer is 15%

you.

(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