Select ALL the correct answers. A supermarket is testing brands of microwave popcorn. For each brand, they recorded the percentage of corn kernels in a package that popped in the recommended cooking time. Let A be the data set of percentages listed below. A = {91, 94, 89, 93, 87, 92, 86, 87, 92, 91} The supermarket then realized they missed a brand. When the new brand was cooked, only 74% of its kernels popped. Let B be the data set of percentages for all 11 packages, including the package where the least kernels popped. Select the statements that are true. The mean of A is the same as the mean of B. Including the percentage for the package where the least kernels popped causes the center of the data to increase. The standard deviation of B is greater

Option C is correct.

The standard deviation of B is greater.

Step-by-step explanation:

Distribution A is given

A = {91, 94, 89, 93, 87, 92, 86, 87, 92, 91}

Then distribution B is all of distribution A including the percentage for the package where the least kernels popped

B = {74, 91, 94, 89, 93, 87, 92, 86, 87, 92, 91}

Taking the statements one at a time

- The mean of A is the same as the mean of B.

Mean = (sum of variables)/(sample size)

It is evident that most of the percentages in distribution A hover around a consistent spot, hence, the mean too will land around that consistent spot. The mean of distribution A as calculated = (902/10) = 90.2

Mean of distribution B should hover around that spot too, but a new addition of 74 is expected to bring the mean down.

Mean of distribution B = (976/11) = 88 73

Hence, statement 1 is incorrect as the mean of A isn't the same as the mean of B.

- Including the percentage for the package where the least kernels popped causes the center of the data to increase.

The center of the data is synonymous with the mean and it is evident that this centre doesn't increase after including the percentage for the package where the least kernels popped causes the center of the data to increase, rather, it drops from 90.2 to 88.73.

Hence, this statement is false.

- The standard deviation of B is greater

Standard deviation measures how far away from the mean the variables of the distribution are. And since we've established that the variables of distribution A hover around a spot, hence, the stamdard deviation is expected to be small. But distribution B has all of the variables of distribution A plus a value that is very far away from the Mean (74), hence, it is easy to see that distribution B will have the greater standard deviation. Let us now calculate.

Standard deviation = σ = √[Σ(x - xbar)²/N]

x = each variable

xbar = mean

N = Sample size

Standard deviation for distribution A = 2.638

Standard deviation for distribution B = 5.293

Hence we can conclude that this statement is true and the standard deviation of distibution B is greater.

Hope this Helps!!!

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