W(x) = −3x− 4; Find w(7)
2 answers:
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Draw ΔABC as shown in the figure below.
m∠B = 137°, b = 14, c = 9.
From the Law of Sines,
sin(C)/c = sin(B)/b
or
sin(C)/9 = sin(137°)/14
sin(C) = (9/14)*sin(137°) = 0.4384
m∠C = arcsin(0.4384) = 26°
Because the sum of angles in a triangle is 180°, therefore
m∠A = 180 -137 - 26 = 17°
From the Law of Sines,
a/sin(A) = b/sin(B)
or
a/sin(17°) = 14/sin(137°)
a = (sin(17°)/sin(127°))*14 = 6
Answer:
The solution for the triangle is
a = 6, b = 14, c = 9
m∠A = 17°, m∠B = 137°, m∠C = 26°
m∠B = 137°, b = 14, c = 9.
From the Law of Sines,
sin(C)/c = sin(B)/b
or
sin(C)/9 = sin(137°)/14
sin(C) = (9/14)*sin(137°) = 0.4384
m∠C = arcsin(0.4384) = 26°
Because the sum of angles in a triangle is 180°, therefore
m∠A = 180 -137 - 26 = 17°
From the Law of Sines,
a/sin(A) = b/sin(B)
or
a/sin(17°) = 14/sin(137°)
a = (sin(17°)/sin(127°))*14 = 6
Answer:
The solution for the triangle is
a = 6, b = 14, c = 9
m∠A = 17°, m∠B = 137°, m∠C = 26°
Solution :
Number of different samples of two technicians possible is = 10, as 2 technicians are to be chosen from a total of 5 technicians.
a). Therefore, 10 different samples are possible of size two.
b).
<u> Sample Shut down Mean</u>
1. Taylor, Hurley 4, 3 3.5
2. Taylor, Gupta 4, 5 4.5
3. Taylor, Rousche 4,6 5
4. Taylor, Huang 4, 3 3.5
5. Hurley, Gupta 3,5 4
6. Hurley, Rousche 3,6 4.5
7. Hurley, Huang 3,3 3
8. Gupta, Rousche 5,6 5.5
9. Gupta, Huang 5,3 4
10. Rousche, Huang 6,3 <u> 4.5 </u>
42
Now mean of sample means = 4.2
Mean of population is = = 4.2
Thus both the means are equal.