Answer:
C
Step-by-step explanation:
Is is a one-tailed test
P(z < 2.45) = 0.9929
1 - 0.9929 = 0.0071
Significant at both 0.05 and 0.01 because both are greater than 0.0071
Step-by-step explanation:
∫₀³⁰ (r/V C₀ e^(-rt/V)) dt
If u = -rt/V, then du = -r/V dt.
∫ -C₀ e^u du
-C₀ ∫ e^u du
-C₀ e^u + C
-C₀ e^(-rt/V) + C
Evaluate between t=0 and t=30.
-C₀ e^(-30r/V) − -C₀ e^(-0r/V)
-C₀ e^(-30r/V) + C₀
C₀ (-e^(-30r/V) + 1)
I got the same answer. Try changing the lowercase v to an uppercase V.
Answer:
Step-by-step explanation:
11.04 = 10(1.02)^n
1.104 = 1.02^n
ln 1.104 = ln 1.02^n
ln 1.104 = n ln 1.02
n = ln 1.104/ ln 1.02
n = 4.99630409516
4.99 can be rounded to 5.
So a reasonable domain would be 0 ≤ x < 5
PART B)
f(0) = 10(1.02)^0
f(0) = 10(1)
f(0) = 10
The y-intercept represents the height of the plant when they began the experiment.
f(1) = 10(1.02)^1
f(1) = 10(1.02)
f(1) = 10.2
(1, 10.2)
f(5) = 10(1.02)^5
f(5) = 10(1.1040808)
f(5) = 11.040808
f(1)=10(1.02)^1
f(1)=10.2
Average rate= (fn2-fn1)/(n2-n1)
=11.04-10.2/(5-1)
=0.22
the average rate of change of the function f(n) from n = 1 to n = 5 is 0.22.
Answer:
25 sq. cm
Step-by-step explanation:
i think so i'm not sure if it is correct
Check the picture below.

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.
![\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%20in%20feet%7D%20%5C%5C%5C%5C%20h%28x%29%20%3D%20-16x%5E2%2Bv_ox%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%26%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%5C%5C%20x%3D%5Ctextit%7Bseconds%7D%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%200%3D-16%281%29%5E2%2B0x%2Bh_o%5Cimplies%200%3D-16%2Bh_o%5Cimplies%2016%3Dh_o%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20h%28x%29%20%3D%20-16x%5E2%2B16~%5Chfill)
quick info:
in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².