19c10x42c4/61c14 what would that equal to?

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!!!

4 0

3 years ago

Population lottery numbers: 56, 42, 19, 8, 24, 13 find the total number of samples of size 4 that can be drawn from this populat

statuscvo [17]

Answer:

The correct answer is there are 15 samples each of size 4 to choose from the population and the population mean is 27.

Step-by-step explanation:

Total number of lottery tickets = 6

Number of samples of size 4 that can be drawn from this population of 6 lottery numbers are = $\left[\begin{array}{ccc}6\\4\end{array}\right]$ = 15.