Answer:
855 cft
Step-by-step explanation:
(22/7)*4*4*17=855cft(rounded )
The total number of possible classifications for the students of this college is found by multiplying 4 (which is the classification for the year level:freshman, sophomore, juniou, senior) and 2 (which is the number of sexes: female and male). So 4 x 2 = 8. There are eight possible classifications, which are:
(Male, Freshman)
(Male, Sophomore)
(Male, Junior)
(Male, Senior)
(Female, Freshman)
(Female, Sophomore)
(Female, Junior)
(Female,Senior)
Answer:
<h3>Add 47.6 and 39.75, then round the answer</h3>
Step-by-step explanation:
If Ramina found the length of two pieces of ribbon to be 47.6 inches and 39.75 inches, the effective strategy of finding the sum of the two lengths is to:
1) First is to add the two values together
47.6 + 39.75
= (47+0.6)+(39+0.75)
= (47+39)+(0.6+0.75)
= 86 + 1.35
= 87.35
2) Round up the answer to nearest whole number.
87.35 ≈ 87 (note that we couldn't round up to 88 because the values after the decimal point wasn't up to 5)
Option C is correct
9514 1404 393
Answer:
(dN)/(dt) = (0.4)/(1200)N(1200-N) -50
142 fish
Step-by-step explanation:
A) The differential equation is modified by adding a -50 fish per year constant term:
(dN)/(dt) = (0.4)/(1200)N(1200-N) -50
__
B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.
(0.4/1200)(N)(1200-N) -50 = 0
N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient
N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square
(N -600)^2 = 210,000 . . . . . simplify
N = 600 + √210,000 ≈ 1058
This steady-state number of fish is ...
1200 - 1058 = 142 . . . . below the original carrying capacity