To solve this problem you must apply the proccedure shown below:
1. You have to r<span>ewrite x=12 in polar form. Then, you have:
12=rCos</span><span>θ
2. Then, you must solve for r, as following:
r=12/Cos</span><span>θ
</span> 3. You have that 1/Cosθ=Sec<span>θ, therefore:
</span> r=12(1/Cos<span>θ)
</span> r=12Sec<span>θ
</span> Therefore, as you can see, the answer is: r=12Secθ<span>
</span>
Answer:
C) The centers will roughly be equal, and the variability of simulation A will be less than the variability of simulation B.
Step-by-step explanation:
The center for Simulation A and Simulation B will be roughly equal.
Overall Sample size of Simulation A = 1500 * 100 = 150000
Overall Sample size of Simulation B = 2000 * 50 = 100000
Since the sample size for Simulation A is greater, the variability of Simulation will be less.
Therefore, The answer is C) The centers will roughly be equal, and the variability of simulation A will be less than the variability of simulation B.
Slope is the rise and run of a point on the graph to the other
the y intercept is where a line crosses the y axis for example, (0,5) so 5 is the y intercept
(0,8), (1,10), (2, 12), (3, 14), (4, 16), (5, 18)
Distance = Speed * Time
D = 25.6 * 4
D = 102.4 miles
In short, Your Final Answer would be: 102.4 miles
Hope this helps!
Answer:
Step-by-step explanation:
Given the differential model
![\dfrac{dP}{dt}=k[M-P(t)]](https://tex.z-dn.net/?f=%5Cdfrac%7BdP%7D%7Bdt%7D%3Dk%5BM-P%28t%29%5D)
We are required to solve the equation for P(t).
