Part a: identity a fraction equivalent to 15/60. show all work.
2 answers:
You might be interested in
(1)
Mean length of all fish in the sample -
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (12 + 5 + 3 + 5 + 8 + 2 + 10 + 9 + 4 + 4)/(10)
= 62/10
=<em> 6.2</em>
(2)
Mean length of adult fish in the sample -
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (12 + 5 + 8 + 10)/(4)
= 35/4
=<em> 8.75</em>
(3)
Mean length of juvenile fish in the sample -
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (5 + 3 + 2 + 9 + 4 + 4)/(6)
= 27/6
<em>= 4.5</em>
(4)
Percentage of sample that were adult fish -
<u>Percentage</u> = (No. of adult fishes)/(Total no. of fishes) × 100
% = (4/10) × 100
<em>% = 40</em>
(5)
Percentage of sample that were juvenile fish -
<u>Percentage</u> = (No. of juvenile fishes)/(Total no. of fishes) × 100
% = (6/10) × 100
<em>% = </em><em>6</em><em>0</em>
(6)
Percentage of sample that were juveniles over 8 inches long -
<u>Percentage</u> = (No. of juveniles over 8 inches)/(Total no. of fishes) × 100
% = (1/10) × 100
<em>% = </em><em>1</em><em>0</em>