<span>quotient of 63 and a number h would be 63/h
14 less than that would be 63 over h - 14
</span>
P(at least 1 missed shot) = 100 - 77.6 = 22.4%
Number of times to expect al least 1 missed shot = 0.224 x 2500 = 560 times.
I would start by multiplying both sides of the inequality by 4 to eliminate the fraction
y + 8 >/= 12
Then subtract 8 from both sides
y >/= 4
Because it is greater than OR equal to, when you graph, you use a solid circle. Greater than means the arrow goes to the right on the line.
So, solid circle on the line on 4, arrow pointing to the right. (Third option from the right)
Step 1:
Calculate the measure of angle ∠ABC
From the triangle in the question,
Step 2:
Calculate the value of AB using the cosine rule below
By substituting the values, we will have
![\begin{gathered} b^2=a^2+c^2-2\times a\times c\times\cos B \\ b^2=10^2+15^2-2\times10\times15\times\cos 115^0 \\ b^2=100+225-300\times(-0.4226) \\ b^2=325+126.78 \\ b^2=451.78 \\ \text{Square root both sides} \\ \sqrt[]{b^2}=\sqrt[]{451.78} \\ b=21.26\operatorname{km} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20b%5E2%3Da%5E2%2Bc%5E2-2%5Ctimes%20a%5Ctimes%20c%5Ctimes%5Ccos%20B%20%5C%5C%20b%5E2%3D10%5E2%2B15%5E2-2%5Ctimes10%5Ctimes15%5Ctimes%5Ccos%20115%5E0%20%5C%5C%20b%5E2%3D100%2B225-300%5Ctimes%28-0.4226%29%20%5C%5C%20b%5E2%3D325%2B126.78%20%5C%5C%20b%5E2%3D451.78%20%5C%5C%20%5Ctext%7BSquare%20root%20both%20sides%7D%20%5C%5C%20%5Csqrt%5B%5D%7Bb%5E2%7D%3D%5Csqrt%5B%5D%7B451.78%7D%20%5C%5C%20b%3D21.26%5Coperatorname%7Bkm%7D%20%5Cend%7Bgathered%7D)
Hence,
The distance of point A to point C is = 21.26km
