Well you can start by drawing a triangle with the information
the size of angle B can be found using the Law of Sine
![\frac{sin(150)}{9.4)} = \frac{sin(B)}{4.8}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bsin%28150%29%7D%7B9.4%29%7D%20%20%3D%20%20%5Cfrac%7Bsin%28B%29%7D%7B4.8%7D%20)
the size of angle C can be found using angle sum of a triangle.
The length of side c can be found using the Law of Cosines
![c^2 = 4.8^2 + 9.4^2 - 2 \times4.8\times9.4\times cos(C)](https://tex.z-dn.net/?f=c%5E2%20%3D%204.8%5E2%20%2B%209.4%5E2%20-%202%20%5Ctimes4.8%5Ctimes9.4%5Ctimes%20%20cos%28C%29)
hope it helps
Answer:
x =(4-√40)/6=(2-√ 10 )/3= -0.387
x =(4+√40)/6=(2+√ 10 )/3= 1.721
Step-by-step explanation: Hope this helped!!!
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Answer:
Step-by-step explanation:
number of cards = 52
number of queen = 4
number of spades = 13
A) probability that the tenth card is a queen
drawn time (r) = 1
position of success(x) = 10th
p = 4/52
P( x,r,p) = ![x-1Cr-1*p^(r) * q^(x-r)](https://tex.z-dn.net/?f=x-1Cr-1%2Ap%5E%28r%29%20%2A%20q%5E%28x-r%29)
p(10,1,4/52) = 9C0(4/52)^1 * (48/52)^9 = 0.0374
B) probability the twentieth card is a spade
x = 20
r = 1
p = 13 / 52
P(20,1,26/52) = 19C0(26/52)^1 * (26/52)^19 = 0.0010
c) The last five cards been spades
p(last five cards been spades )
p(48..52, 5, 13/52 ) = 47...52C4(13/52)^5 * (39/52)^48..52 - 5
Answer:
C. Use the size of the population as a parameter in the operating characteristics formulas.
Step-by-step explanation:
Models with a finite calling population use the size of the population as a parameter in the operating characteristics formulas.