The Wildcats and the Leopards are evenly matched football teams. When they play, there is a 0.5 probability that the Wildcats will win. If they play 9 times, what is the probability that the Wildcats will win 4 of the games? Round your answer to the nearest tenth of a percent O A. 7.0% OB. 16.4% O C. 24.6% O D. 0.2%​

15.36 per mile

but forgot to round

i am pretty sure

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

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.