In the 9th and 10th grades at Jefferson High School, there are 236 students. Of thoes students, 121 are in 9th grade, Of the 214
students who are right- handed, 103 of them are in 10th grade.
2 answers:
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The smaller the angle subtended by a side, the smaller the length of the
side.
The correct responses are;
Question 1: The list of sides from shortest to longest are;
- MO/Shortest MO/Medium and MO/Longest
a) <u>Friday</u>
b) <u>70 minutes</u>
c) <u>40%</u>
d) Yes<u>,</u> <u>the sum of the </u><u>mean</u><u> number of </u><u>minutes spent</u><u> on </u><u>aerobic</u><u> training and the mean number of minutes spent on </u><u>strength</u><u> training is equal to the mean </u><u>total</u><u> number of minutes spent </u><u>training.</u>
From the given diagram, we have, the measure of the third angle, ∠O, is
found as follows;
∠O = 180° - 54° - 61° = 65°
Therefore, ∠O = The largest angle
We get;
The longest side is opposite the largest angle, which gives;
The shortest side is the side opposite ∠N (54°)=
The next shortest side is the side opposite ∠M(61°) =
The longest side is the side opposite ∠O(65°) =
a) The time spent training on Tuesday = 60 + 10 = 70 minutes
The time spent training on Thursday = 50 + 30 = 80 minutes
The time spent training on Friday = 45 + 40 = 85 minutes
Therefore, the day the athlete spent the longest total amount of time training is on <u>Friday</u>
b) The time spent training on Monday = 10 + 20 = 30 minutes
The time spent training on Wednesday = 20 + 15 = 35 minutes
Therefore, we get;
30, 35, 70, 80, and 85
The median total number of minutes the athlete spent training each day = <u>70 minutes</u>
<u />
c) The time spent strength training = 20 + 10 + 15 + 30 + 45 = 120
The total number of minutes the athlete spent training = 70 + 80 + 85 + 30 + 35 = 300
=
×
%
d) The mean number of minutes spent on strength training is found as follows;
The mean number of minutes spent on aerobic training is found as follows;
The mean total number of minutes spent training, =
= 60
Therefore;
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