Functions A,B and C are linear functions. Some values of function A are shown in the table. Function B has y-intercept of (3,0)
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The time intervals when the riders could see Niagara falls are; 0.834 < t < 1.416 and (3.084, 3.666)
<h3>How to interpret Cycle Graphs?</h3>
From the diagram attached, we can say that;
Period = 2π/k
where;
k = 2π/2.25
k = 8π/9
Thus;
h(t) = -(48/2) cos (8π/9)t + ((48/2) + 0.5)
h(t) = -24cos (8π/9)t + 24.5
Riders can see Niagara falls if they are higher than 41 meters above the ground. Thus;
41 = -24cos (8π/9)t + 24.5
41 - 24.5 = -24cos (8π/9)t
16.5 = -24cos (8π/9)t
-0.6875 = cos (8π/9)t
cos⁻¹0.6875 = (8π/9)t
t = 0.834 min
Thus, time interval is between;
0.834 < t < (2.25 - 0.834)
⇒ 0.834 < t < 1.416 and
(2.25 + 0.834) < t < (2.25 + 1.416)
⇒ (3.084, 3.666)
Read more about Cycle Graphs at; brainly.com/question/24461724
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from what I know
A=4
B=3
C=2
and a plus adds 0.33 and a minus minuses 0.33
question wants us to find yearly average or average per 1 year
since these grades span over 2 years, we must divide the total average by 2
averagefresh=sum of fresh grades/4 grades per year
averagesoph=sum of soph grades/4 grades per year
yearlyaverage=(averagefresh+averagesoph)/2
freshman:
(C)+(B)+(A)+(C-)=2+3+4+(2-0.33)=10.67
sophomore:
(B+)+(A-)+(C+)+(B-)=(3+0.33)+(4-0.33)+(2+0.33)+(3-0.33)=12
averagefresh=10.67/4=2.6675
averagesoph=12/4=3
yearlyaverage=(2.6675+3)/2=2.83375
so it's about a average