Answer:
The second time when Luiza reaches a height of 1.2 m = 2 08 s
Step-by-step explanation:
Complete Question
Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.
H(t) = -0.6 cos (2pi/2.5)t + 1.5.
What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.
Solution
Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as
H(t) = -0.6cos(2π/2.5)t + 1.5
What is t when H = 1.2 m
1.2 = -0.6cos(2π/2.5)t + 1.5
0.6cos(2π/2.5)t = 1.2 - 1.5 = -0.3
Cos (2π/2.5)t = (0.3/0.6) = 0.5
Note that in radians,
Cos (π/3) = 0.5
This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,
Cos (5π/3) = 0.5
So,
Cos (2π/2.5)t = Cos (5π/3)
(2π/2.5)t = (5π/3)
(2/2.5) × t = (5/3)
t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.
Hope this Helps!!!
Answer:
c) 62.
Step-by-step explanation:
Let x represent number of customers came in on the fifth day.
We have been given that the mean number of customers for those four days is 52. So total number of customers on 4 days would be 4 times 52 that is 208 customers.
We know that mean of a data set is equal to sum of all data points divided by number of data points.
Total number of customers on 5 days would be 
and total number of days is 5.
Since we need a mean of 54, so we can set mean of 5 days equal to 54 as:
Multiply both sides by 5:
Therefore, 62 customers must come in on the fifth day to make the five-day mean 54 and option 'c' is the correct choice.
It should be B because it’s going down 4
