Answer:
y = 6cos(π/2)t
Step-by-step explanation:
Given in the question that,
in a simple harmonic motion in which
at t=0 the amplitude is 6 cm, so the simple harmonic motion will be in cosine form. As in case of sine function, the value of function at t=0 is 0.
And,
the period is given as 4 sec
We know that model equation of simple harmonic motion is
<h3>y = a. cosb(t + c) + d</h3>
here,
a is amplitude
c is Horizontal shift or phase shift
d is Vertical shift
<h3>Step1</h3><h3>Find b</h3>
Period = 2 π / b
4 = 2π / b
b = 2π/4
b = 1π/2
<h3>Step2</h3><h3>Find shifts</h3>
Horizontal and vertical shifts are 0.
So, c = 0 and d = 0
<h3>Step3</h3><h3>Plug those values in the above in the equation</h3>
y = 6cos π/2 (t + 0) + 0
y = 6cos(π/2)t
Answer:
x + 3y = 5
x + 3y = 5
Thus the equations are equal and they are consistent, graph is coinciding , and they have infinite solutions
Answer:
x = 45
Step-by-step explanation:
0.36(18) + 0.04x = 0.12(24 + x)
6.48 + 0.04x = 2.88 + 0.12x
0.04x - 0.12x = 2.88 - 6.48
-0.08x = -3.6
0.08x = 3.6
x = 3.6/0.08
x = 45
Answer:
0.47 is the probability it rains at least 2 out of any randomly selected 5 days during the given time of year
Step-by-step explanation:
We are given the following information:
We treat training as a success.
P(Rain) = 30% = 0.30
Then the chances of rain on each day follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 5
We have to evaluate:
0.47 is the probability it rains at least 2 out of any randomly selected 5 days during the given time of year
The answer is the substitution property
