A) For this problem, we will need to use a normal calculation, in that we find the z-score and the area to the right using Table A.
z = (10 - 7.65) / 1.45
z = 1.62
area to the left for a z-score of 1.62 = 0.9474
area to the right for a z-score of 1.62 = 0.0526
The probability that a randomly selected ornament will cost more than $10 is 0.0526 or 5.26%.
B) For this problem, we will use the binomial probability formula since the problem is asking for the probability that exactly 3 ornaments cost over $10. There are two forms of this equation. One is <em>nCr x p^r x q^n-r</em> and the other is <em>(n r) x p^r x (1 - p)^n-r</em>. I will show both formulas below.
8C3 x 0.0526^3 x 0.9474^5
(8 3) x 0.0526^3 x 0.9474^5
With both equations, the answer is the same. Whichever you are more familiar or comfortable with is the one I would recommend you use.
The probability that exactly 3 of the 8 ornaments cost over $10 is 0.00622 or 0.622%.
Hope this helps!! :)
Y = -3x + 5
5x - 4y = -3
5x - 4(-3x + 5) = -3
5x + 4(3x) - 4(5) = -3
5x + 12x - 20 = -3
17x - 20 = -3
+ 20 + 20
17x = 17
17 17
x = 1
y = -3x + 5
y = -3(1) + 5
y = -3 + 5
y = 2
(x, y) = (1, 2)
A8% is X
10% is 700-X
X(.08)+ .10(700-X)= 64
.08x+ 70-.10X=64
-.02X+70=64
-70=-70
-.02x= -6
Divide by -.02
X= 300
Answer $300 was invested at 8%
Check 300 times .08= 24. 700-300=400 times .10 = 40 and 24+40=64
We are going to take out the dividends step by step:
Dell:
(66) * (1.79) = 118.14 $
Coca Cola:
(95) * (2.62) = 248.9 $
Nike:
(180) * (1.18) = 212.4 $
Adding the three dividends we have:
118.14 + 248.9 + 212.4 = 579.44 $
Answer:
Manuel does receive 579.44 $ in dividends every year
