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Norma-Jean [14]
3 years ago
6

The following two-way table shows the data for the students of two different grades in a school:

Mathematics
2 answers:
Alenkasestr [34]3 years ago
8 0
There are 18 students in Grade 7, 10 are in band, so 10/18 = 0.56 are in band.

There are 12 students in Grade 8, 10 are in band, so 10/12 = 0.83 are in band.

Students in Grade 8 are more likely to be members of the band.

bearhunter [10]3 years ago
8 0
<h2>Answer:</h2>

Based on the relative row frequency, the students of Grade 8 are more likely to be members of the school band.

<h2>Step-by-step explanation:</h2>

The relative row frequency is obtained by dividing each entry of by the corresponding row total.

i.e. the first row is obtained by dividing each entry of the first row by the row total.

Similar process holds for second row and third row.

Here on obtaining the  relative row frequency we have:

The relative row frequency that a Grade 7 student like to be a member of school band= 0.56

and the relative row frequency that a Grade 8 student like to be a member of school band= 0.83

          Since,   0.83>0.56

Hence, we have:

Grade 8 student is likely to be a member of school band.

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A spinner with 3 sectors is shown in the figure below. The measures of the central angles of the red, white, and blue sectors, i
RideAnS [48]

Given that a spinner with 3 sectors. The measures of the central angles of the red, white, and blue sectors, in that order, are in the ratio 2:3:4.

So we can write:

Measure of the central angle of the red sector = 2x

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Measure of the central angle of the blue sector = 4x


We know that sum of central angle of the circle is 360 degree so the sum of above three angles will also be 360 degree

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Then measure of the central angle of the red sector = 2x=2*40= 80 degree

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We have to find the probability for red sector so first we need to find are of the red sector which is given by formula :

Area = \frac{\pi r^2 \theta}{360} , where r is the radius of the spinner.

Area = \frac{\pi r^2 *80}{360}

Area = \frac{2 \pi r^2}{9}

Area of the spinner is circular so that is given by

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Now the  probability that the arrow stops in the red sector is given by ratio or area of red sector and the area of spinner

Probability = \frac{\frac{2}{9}\pi r^2}{\pi r^2}

Probability = \frac{\frac{2}{9}}{1}

Probability = \frac{2}{9}

Hence required probability is 2/9.

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