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kkurt
1 month ago
12

Twelve of the 20 students in Mr. Skinner’s class brought lunch from home. Fourteen of the 21 students in Ms. Cho’s class brought

lunch from home. Siloni is using two 15-section spinners to simulate randomly selecting students from each class and predicting whether they brought lunch from home or will buy lunch in the cafeteria.
If each spinner is divided into 15 congruent sectors, how does the spinner representing Mr. Skinner’s class differ from the spinner representing Ms. Cho’s class?
One more sector of the Skinner-class spinner will represent bringing lunch from home.
One fewer sector of the Skinner-class spinner will represent bringing lunch from home.
Two more sectors of the Skinner-class spinner will represent bringing lunch from home.
Two fewer sectors of the Skinner-class spinner will represent bringing lunch from home.
Mathematics
2 answers:
Svet_ta [12.7K]1 month ago
9 0

Solution:

In Mr. Skinner's class, the count of students bringing lunch from home is 12 out of 20.

Fraction of students who brought lunch from home in Mr. Skinner's class=\frac{12}{20}=\frac{3}{5}

For Ms. Cho's class, the number who brought lunch from home is 14 out of 21.

Fraction of students who brought lunch from home in Ms. Cho's class=\frac{14}{21}=\frac{2}{3}

Siloni is utilizing two spinners with 15 equal sections to randomly select students from the classes and predict whether they brought lunch or will purchase it from the cafeteria.

Number of Equal sections in each Spinner=15

To visualize the students from Mr. Skinner's class who brought lunch using a Spinner with 15 equal sections =\frac{9}{15}

For Ms. Cho's class, using a Spinner with 15 equal sections =\frac{10}{15}

Mr. Skinner's Class +1 = Ms. Cho's Class

This means that the spinner for Ms. Cho's class will include one additional section representing students who brought lunch.

Option A signifies that one additional section on Mr. Skinner's spinner represents students who brought lunch, reflecting Ms. Cho's class.

AnnZ [12.3K]1 month ago
7 0

Response: A

Detailed explanation:

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n=110, \sum x_i y_i = \sum (X-\bar X)(Y-\bar Y) =7625.9,\sum x^2_i=\sum (x-\bar x)^2 =1248.9, sum y^2_i=\sum(y-\bar y)^2 =94228.8

\sum Y_i =17375, \sum X_i = 7665.5

Part a

The slope can be calculated using this formula:

\hat \beta_1 =\frac{\sum (x-\bar x) (y-\bar y)}{\sum (x-\bar x )^2}

Following the substitutions, we have:

\hat \beta_1 =\frac{7625.9}{1248.9}=6.106

The intercept can be determined with this formula:

\hat \beta_o = \bar y -\hat \beta_1 \bar x

Average values for x and y can be calculated this way:

\bar X=7665.5/110 =69.686, \bar y= 17375/110=157.955

Replacing yields:

\hat \beta_o = 157.955 -6.106 (69.686)=-267.548

Part b

The correlation coefficient can be calculated using the following formula:

r=\frac{\sum (x-\bar x)(y-\bar y) }{\sqrt{[\sum (x-\bar x)^2][\sum(y-\bar y)^2]}}

In our situation:

n=110, \sum x_i y_i = \sum (X-\bar X)(Y-\bar Y) =7625.9,\sum x^2_i=\sum (x-\bar x)^2 =1248.9, sum y^2_i=\sum(y-\bar y)^2 =94228.8

We can compute the correlation coefficient by substituting values:

r=\frac{7625.9}{\sqrt{[1248.9][94228.8]}}=0.657

The coefficient of determination is r^2 = 0.657^2 =0.432

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