Answer:
183$
Step-by-step explanation:
.........................
Answer:
<h2><em>
2x-4</em></h2>
Step-by-step explanation:
Area of a rectangle = Length * Width
Given parameters
Area A = 8x2 – 10x – 12
Length of the rectangle = 4x+3
Required
Width of the rectangle.
Substituting the given parameters into the formula
8x2 – 10x – 12 = (4x+3)*width
width = 8x2 – 10x – 12
/4x+3
S
Factorizing the numerator
8x² – 10x – 12
= 2(4x²-5x-6)
= 2(4x²-8x+3x-6)
= 2(4x(x-2)+3(x-2))
= 2(4x+3)(x-2)
Width = 2(4x+3)(x-2)/4x+3
Width = 2(x-2)
Width = 2x-4
<em>Hence the width of the rectangle is 2x-4</em>
Answer:
M(h) = 73.18025h
Step-by-step explanation:
The composite function is ...
M(B(L(h))) = M(B(28.75h)) = M(1.78(28.75h)) = M(51.175h)
= 1.43(51.175h) = 73.18025h
The composite function is ...
M(h) = M(B(L(h))) = 73.18025h
Answer:
a) P=0.0225
b) P=0.057375
Step-by-step explanation:
From exercise we have that 15% of items produced are defective, we conclude that probabiity:
P=15/100
P=0.15
a) We calculate the probabiity that two items are defective:
P= 0.15 · 0.15
P=0.0225
b) We calculate the probabiity that two of three items are defective:
P={3}_C_{2} · 0.85 · 0.15 · 0.15
P=\frac{3!}{2!(3-2)!} · 0.019125
P=3 · 0.019125
P=0.057375
Answer:
0.362
Step-by-step explanation:
When drawing randomly from the 1st and 2nd urn, 4 case scenarios may happen:
- Red ball is drawn from the 1st urn with a probability of 9/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this case to happen is (9/10)*(1/6) = 9/60 = 3/20 or 0.15. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 5 blue)/(8 red + 1 blue + 5 blue) = 6/14 = 3/7.
- Red ball is drawn from the 1st urn with a probability of 9/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (9/10)*(5/6) = 45/60 = 3/4 or 0.75. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 4 blue)/(8 red + 1 blue + 1 red + 4 blue) = 5/14
- Blue ball is drawn from the 1st urn with a probability of 1/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (1/10)*(5/6) = 5/60 = 1/12. The probability that a ball drawn randomly from the third urn is blue given this scenario is (4 blue)/(9 red + 1 red + 4 blue) = 4/14 = 2/7
- Blue ball is drawn from the 1st urn with a probability of 1/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this event to happen is (1/10)*(1/6) = 1/60. The probability that a ball drawn randomly from the third urn is blue given this scenario is (5 blue)/(9 red + 5 blue) = 5/14.
Overall, the total probability that a ball drawn randomly from the third urn is blue is the sum of product of each scenario to happen with their respective given probability
P = 0.15(3/7) + 0.75(5/14) + (1/12)*(2/7) + (1/60)*(5/14) = 0.362
