The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 49 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.4 pounds, what is the probability that the sample mean will be each of the following? Appendix A Statistical Tables a. More than 58 pounds b. More than 57 pounds c. Between 55 and 57 pounds d. Less than 55 pounds e. Less than 47 pounds

the value of x is 9 ⇒ answer b

step-by-step explanation:

* we can solve this problem using cosine and sine rule

- in δ abc

∵ ab = 36 , bc = 28 , ac = 16

- lets find the measure of angle b using the cosine rule

∵ cos b = [ab² + bc² - ac²]/2(ab)(bc)

∴ cos b = [36² + 28² - 16²]/2(36)(28)

∴ cos b = 19/21

∴ m∠b = cos^-1(19/21) ≅ 25°

- lets find the measure of angle a using the sine rule

∵ sina/bc = sinb/ac = sinc/ab

∴ sina/28 = sin25°/16 ⇒ by using cross multiplication

∴ sina = 28 × sin25°/16

∴ sina = 0.7396

∴ m∠a = sin^-1(0.7396) ≅ 48°

- from the figure m∠abd = m∠dbc

∵ m∠abc = 25°

∴ m∠abd = 25°/2 = 12.5°

- in δabd

∵ m∠a = 48° , m∠abd = 12.5°

∵ the sum of the measures of the interior angles of any δ is 180°

∴ m∠adb = 180° - (48° + 12.5°) = 119.5°

∵ ab = 36

∵ ad = x

- use the sine rule to find x

∵ sin∠adb/ab = sin∠abd/x

∴ sin119.5°/36 = sin12.5°/x ⇒ by using cross multiplication

∴ x = 36 × sin12.5°/sin119.5° = 8.95 ≅ 9

* the value of x is 9

vhhhbnbb. ddben jenner v

134.6 us dollars

step-by-step explanation:

1 us dollar = 1.33 canadian dollars

so, 150 us dollars = 150 × 1.33 = $199.5.

in canada, the money mitchell spent = 20 canadian dollars.

remaining canadian dollars with him = 199.5 - 20 = 179.5.

after mitchell returned to us, he exchanged the remaining canadian dollars to us dollars.

179.5 canadian dollars = 179.5 × 0.75 = 134.6 us dollars.

not sure srry man