The table below shows the weights of 8 different bears at a zoo. Type of Bear Weight (pounds)
Asiatic Black Bear 225
Black Bear 300
Brown Bear 550
Panda Bear 200
Polar Bear 1,000
Sloth Bear 300
Spectacled Bear 280
Sun Bear 100
If the weight of the polar bear is removed, which statement is true?
A The mean decreases more than the median because the polar bear is a high outlier.
B The mean decreases less than the median because the polar bear is a high outlier.
C The mean decreases more than the median because the high value balances the low value.
D The mean decreases less than the median because the high value balances the low value.​

for future students the answer is b

hope i

-lavira: )

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