The height at time t is
h(t) = 144 - 16t²
When t = 0, then h = 144.
Therefore the height from the ground is 144 when the object is dropped.
When the object reaches the ground, h = 0.
Therefore
144 - 16t² = 0
t² = 144/16 = 9
t = 3 s
Answer:
The object reaches the ground in 3 seconds.
<em>Answer:</em>
<em>There would be 173,535 lionfish after 6 years.</em>
<em>Step-by-step explanation:</em>
<em>Since lionfish are considered an invasive species, with an annual growth rate of 67%, ya scientist estimates there are 8,000 lionfish in a certain bay after the first year, A) to write the explicit equation for f (n) that represents the number of lionfish in the bay after n years; B) determine how many lionfish will be in the bay after 6 years; and C) if scientists remove 1,200 fish per year from the bay after the first year, determine what is the recursive equation for f (n); the following calculations must be performed:</em>
<em></em>
<em>A)</em>
<em>8000 x 1.67 ^ n = f </em>
<em>B)</em>
<em>8000 x 1.67 ^ 6 = X</em>
<em>8000 x 21.691961596369 = X</em>
<em>173,535.692770952 = X </em>
<em>C)</em>
<em>(8000 - 1200 x 1 ^ n) x 1.67 ^ n = f</em>
<em>Therefore, there would be 173,535 lionfish after 6 years.</em>
if the diameter is 20, the its radius must be half that or 10.
![\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20sector%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20r%5E2%7D%7B360%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20%5Ctheta%20%3D%5Cstackrel%7Bdegrees%7D%7Bangle%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20A%3D5%5Cpi%20%5C%5C%20r%3D10%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%205%5Cpi%20%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20%2810%29%5E2%7D%7B360%7D%5Cimplies%205%5Cpi%20%3D%5Ccfrac%7B5%5Cpi%20%5Ctheta%20%7D%7B18%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B5%5Cpi%20%7D%7B5%5Cpi%20%7D%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%201%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%2018%3D%5Ctheta%20%5Cend%7Barray%7D)
Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when 
. So
Rouding up
We need a sample size of at least 719
