Compare the different types of random sampling methods. Describe examples in which stratified sampling and cluster sampling should be used. Do not cite examples already provided by peers. In replies to peers, provide analysis and recommendations.

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

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the given growth  function for the length of the fish (cm) is

f(m) = 5(1.07)m = 5.35m

where 

m = months

at the end of the study, the length is 9.19 cm. the  time for conducting the study is determined as

5.35m = 9.19

m = 1.7178 months

part a.

the study was performed over a time period  of about 1.72 months. therefore  the domain for  plotting the function is 0 < = m < = 1.72.

note:

if growth  continues at this rate for up to m months, then the domain would be 0< =m< =m.

part b.

the growth function is a linear function.

when m=0, f=0.

therefore the initial growth is zero, which represents the y-intercept.

part c.

assume that the growth function is valid for 0< =m< =9.

the average rate of change of the function is the slope of the linear function, which is 5.35 cm/month.