Answer:
160m/s
Step-by-step explanation:
The object can hit the ground when t = a; meaning that s(a) = s(t) = 0
So, 0 = -16a² + 400
16a² = 400
a² = 25
a = √25
a = 5 (positive 5 only because that's the only physical solution)
The instantaneous velocity is
v(a) = lim(t->a) [s(t) - s(a)]/[t-a)
Where s(t) = -16t² + 400
and s(a) = -16a² + 400
v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)
v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)
v(a) = lim (t->a) -16(t² - a²)(t-a)
v(a) = -16lim t->a (t²-a²)(t-a)
v(a) = -16lim t->a (t-a)(t+a)/(t-a)
v(a) = -16lim t->a (t+a)
But a = t
So, we have
v(a) = -16lim t->a 2a
v(a) = -32lim t->a (a)
v(a) = -32 * 5
v(a) = -160
Velocity = 160m/s
Answer:
The probability that it will take more than 10 minutes for the next student to arrive at the library parking lot is 0.0821.
Step-by-step explanation:
The random variable <em>X</em> is defined as the amount of time until the next student will arrive in the library parking lot at the university.
The random variable <em>X</em> follows an Exponential distribution with mean, <em>μ</em> = 4 minutes.
The probability density function of <em>X</em> is:

The parameter of the exponential distribution is:

Compute the value of P (X > 10) as follows:


Thus, the probability that it will take more than 10 minutes for the next student to arrive at the library parking lot is 0.0821.
d ( f o c) / dt = 9
( solution is attached below)
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