Answer:
a) Figure attached
b) ![v_r = \sqrt{3.55^2 +10.94^2}=11.50 mph](https://tex.z-dn.net/?f=%20v_r%20%3D%20%5Csqrt%7B3.55%5E2%20%2B10.94%5E2%7D%3D11.50%20mph)
![\theta = tan^{-1} (\frac{10.94}{3.55})=72.02](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B10.94%7D%7B3.55%7D%29%3D72.02)
Step-by-step explanation:
Part a
See the figure attached.
Part b
For this case first we need to find the vectors of velocity for the boat and the wind like this:
![b= = mph](https://tex.z-dn.net/?f=%20b%3D%20%3C10%20cos%2865%29%2C%2010%20sin%2865%29%3E%3D%20%3C4.23%2C%209.06%3E%20mph)
![w= = mph](https://tex.z-dn.net/?f=%20w%3D%20%3C2%20cos%28110%29%2C%202%20sin%28110%29%3E%3D%20%3C-0.68%2C%201.88%3E%20mph)
And now if we want to find the resulting velocity we just need to add the vector:
![b + w = =mph](https://tex.z-dn.net/?f=%20b%20%2B%20w%20%3D%20%3C4.23-0.68%2C%209.06%2B1.88%3E%3D%3C3.55%2C%2010.94%3Emph)
And the resultant magnitude would be:
![v_r = \sqrt{3.55^2 +10.94^2}=11.50 mph](https://tex.z-dn.net/?f=%20v_r%20%3D%20%5Csqrt%7B3.55%5E2%20%2B10.94%5E2%7D%3D11.50%20mph)
And if we want the resultant angle we can do this:
![\theta = tan^{-1} (\frac{10.94}{3.55})=72.02](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B10.94%7D%7B3.55%7D%29%3D72.02)
Answer:
51
Step-by-step explanation:
The order goes as follows:
Lower Extreme, Lower Quartile, Median, Higher Quartile, Higher Extreme
The Higher Extreme is the highest quantity.
I am joyous to assist you anytime.
Answer:
3550 ft.
Step-by-step explanation:
y = 500(5) + 1050
= 2500 + 1050
= 3550 ft,
1. certain
2. unlikely
3.impossible
4.neither likely of unlikely