If one is to choose among given choices and the order is not important, we use the concept of combination. First, we calculate for the sample space or number available since there is a total number of 12 electives and a student may choose 2 out of them.
S = 12C2
That is "the sample space is equal to combination of 12 taken 2". The answer to this is equal to 66.
Next, we determine the number of outcomes. The equation will be,
O = (5C1) x (3C1)
That is "outcome is equal to combination of 5 taken 1 times combination of 3 taken 1". This is equal to 15. The probability is equal to,
P = O/S
Substituting,
P = (15/66) = 0.227270
The answer to this item is the third choice.
0.08(y + -1) + 0.12y = 0.14 + -0.05(10)
Reorder the terms:
0.08(-1 + y) + 0.12y = 0.14 + -0.05(10)
(-1 * 0.08 + y * 0.08) + 0.12y = 0.14 + -0.05(10)
(-0.08 + 0.08y) + 0.12y = 0.14 + -0.05(10)
Combine like terms: 0.08y + 0.12y = 0.2y
-0.08 + 0.2y = 0.14 + -0.05(10)
Multiply -0.05 * 10
-0.08 + 0.2y = 0.14 + -0.5
Combine like terms: 0.14 + -0.5 = -0.36
-0.08 + 0.2y = -0.36
Solving
-0.08 + 0.2y = -0.36
Solving for variable 'y'.
Move all terms containing y to the left, all other terms to the right.
Add '0.08' to each side of the equation.
-0.08 + 0.08 + 0.2y = -0.36 + 0.08
Combine like terms: -0.08 + 0.08 = 0.00
0.00 + 0.2y = -0.36 + 0.08
0.2y = -0.36 + 0.08
Combine like terms: -0.36 + 0.08 = -0.28
0.2y = -0.28
Divide each side by '0.2'.
y = -1.4
Simplifying
y = -1.4
Answer:
Step-by-step explanation:
v = {[(20sin36°)i + (20cos36°)j] + 10i} mi/h
vE = 20sin36º + 10 = 21.76 mi/h
vN = 20cos36° = 16.18 mi/h
v = √(vE2 + vN2) = √(21.762 + 16.182) mi/h = 27.12 mi/h
θ = tan-1(vN/vE) = tan-1(16.18/21.76) = 36.6º north of east
Width = x
Length = x+18
Assuming the table is rectangular:
Area = x(x + 18)
Therefore:
x(x + 18) <span>≤ 175
x^2 + 18x </span><span>≤ 175
Using completing the square method:
x^2 + 18x + 81 </span><span>≤ 175 + 81
(x + 9)^2 </span><span>≤ 256
|x + 9| </span><span>≤ sqrt(256)
|x + 9| </span><span>≤ +-16
-16 </span>≤ x + 9 <span>≤ 16
</span>-16 - 9 ≤ x <span>≤ 16 - 9
</span>-25 ≤ x <span>≤ 7
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But x > 0 (there are no negative measurements):
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Therefore, the interval 0 < x </span><span>≤ 7 represents the possible widths.</span><span>
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