<h2>a.</h2><h3>Solution:</h3>








<h3>Answer:</h3>
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<h2>b.</h2><h3>Solution:</h3>




<h3>Answer:</h3>
<u>T</u><u>h</u><u>e</u><u> </u><u>f</u><u>a</u><u>m</u><u>i</u><u>l</u><u>y</u><u> </u><u>d</u><u>i</u><u>d</u><u> </u><u>s</u><u>p</u><u>e</u><u>n</u><u>d</u><u> </u><u>$</u><u>4</u><u>4</u><u>.</u><u>1</u><u>0</u><u> </u><u>o</u><u>n</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>m</u><u>e</u><u>a</u><u>l</u><u> </u><u>i</u><u>n</u><u> </u><u>t</u><u>o</u><u>t</u><u>a</u><u>l</u><u>.</u>
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Answer:
20 students should be on each float
Step-by-step explanation:
Number of seniors = 100
Number of juniors = 80
To find number of students that should be on each float, find highest common factor (H.C.F) of 100 and 80
Write prime factorisation of 100 and 80.
100 =
× 
80 =
× 5
So,
H.C.F(100, 80) =
× 5 = 4 × 5 = 20
Therefore,
20 students should be on each float
214.3125
Happy to help (Mark brainliest)
Answer:
.
Step-by-step explanation:
DC = 16 and
DB = 30 so
CB = 14
DB = 30 and
EB = 49 so
EB = 19