What is the equation of the line that passes through the point (-2,5) and has a slope of -4?
1 answer:
Answer:
Step-by-step explanation:
We are asked to find the equation of a line given a point and the slope. We will use the point-slope formula.
In this formula, m is the slope and (x₁, y₁) is the point the line passes through. The slope of this line is -4 and the line passes through (-2,5).
- m= -4
- x₁= -2
- y₁ = 5
Substitute the values into the formula.
We are finding the equation of the line, so we should find the slope-intercept form or y=mx+b ( where m is the slope and b is the y-intercept).We must isolate the variable y on one side of the equation.
First, distribute the -4 on the right side of the equation. Multiply each term inside the parentheses by -4.
5 is being subtracted from y. The inverse operation of subtraction is addition, so we will add 5 to both sides of the equation.
The equation of the line is y= -4x -3. The slope of the line is -4 and the y-intercept is -3.
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Answer:
Step-by-step explanation:
Given the following information:
![\begin{tabular} {|p{1.5cm}|p{1.5cm}|p{1.2cm}|p{1.2cm}|p{1.2cm}|} \multicolumn{1}{|p{1.5cm}|}{State of economy}\multicolumn{1}{|p{2.6cm}|}{Probability of State of economy}\multicolumn{3}{|p{4.8cm}|}{Rate of Return if State Occurs}\\[1ex] \multicolumn{1}{|p{1.5cm}|}{}\multicolumn{1}{|p{2.6cm}|}{}\multicolumn{1}{|c|}{Stock A}&StockB&Stock C\\[2ex] \multicolumn{1}{|p{1.5cm}|}{Boom}\multicolumn{1}{|p{2.6cm}|}{0.66}\multicolumn{1}{|p{1.27cm}|}{0.09}&0.03&0.34\\ \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cp%7B1.5cm%7D%7Cp%7B1.5cm%7D%7Cp%7B1.2cm%7D%7Cp%7B1.2cm%7D%7Cp%7B1.2cm%7D%7C%7D%0A%5Cmulticolumn%7B1%7D%7B%7Cp%7B1.5cm%7D%7C%7D%7BState%20of%20economy%7D%5Cmulticolumn%7B1%7D%7B%7Cp%7B2.6cm%7D%7C%7D%7BProbability%20of%20State%20of%20economy%7D%5Cmulticolumn%7B3%7D%7B%7Cp%7B4.8cm%7D%7C%7D%7BRate%20of%20Return%20if%20State%20Occurs%7D%5C%5C%5B1ex%5D%20%0A%5Cmulticolumn%7B1%7D%7B%7Cp%7B1.5cm%7D%7C%7D%7B%7D%5Cmulticolumn%7B1%7D%7B%7Cp%7B2.6cm%7D%7C%7D%7B%7D%5Cmulticolumn%7B1%7D%7B%7Cc%7C%7D%7BStock%20A%7D%26StockB%26Stock%20C%5C%5C%5B2ex%5D%0A%5Cmulticolumn%7B1%7D%7B%7Cp%7B1.5cm%7D%7C%7D%7BBoom%7D%5Cmulticolumn%7B1%7D%7B%7Cp%7B2.6cm%7D%7C%7D%7B0.66%7D%5Cmulticolumn%7B1%7D%7B%7Cp%7B1.27cm%7D%7C%7D%7B0.09%7D%260.03%260.34%5C%5C%0A%5Cend%7Btabular%7D)
Part A:
The expected return on an equally weighted portfolio of these three stocks is given by:
![0.66[0.33 (0.09) + 0.33 (0.03) + 0.33(0.34)] \\ +0.34[0.33 (0.23) + 0.33(0.29) +0.33(-0.14)] \\ \\ =0.66(0.0297 + 0.0099 + 0.1122)+0.34(0.0759+0.0957-0.0462) \\ \\ =0.66(0.1518)+0.34(0.1254)=0.1002+0.0426=0.1428=\bold{14.28\%}](https://tex.z-dn.net/?f=0.66%5B0.33%20%280.09%29%20%2B%200.33%20%280.03%29%20%2B%200.33%280.34%29%5D%20%5C%5C%20%2B0.34%5B0.33%20%280.23%29%20%2B%200.33%280.29%29%20%2B0.33%28-0.14%29%5D%20%5C%5C%20%20%5C%5C%20%3D0.66%280.0297%20%2B%200.0099%20%2B%200.1122%29%2B0.34%280.0759%2B0.0957-0.0462%29%20%5C%5C%20%20%5C%5C%20%3D0.66%280.1518%29%2B0.34%280.1254%29%3D0.1002%2B0.0426%3D0.1428%3D%5Cbold%7B14.28%5C%25%7D)
Part B:
Value of a portfolio invested 21 percent each in A and B and 58 percent in C is given by
For boom: 0.21(0.09) + 0.21(0.03) + 0.58(0.34) = 0.0189 + 0.0063 + 0.1972 = 0.2224 or 22.24%.
For bust: = 0.21(0.23) + 0.21(0.29) + 0.58(-0.14) = 0.0483 + 0.0609 - 0.0812 = 0.028 or 2.8%
Expected return = 0.66(0.2224) + 0.34(0.028) = 0.1468 + 0.00952 = 0.1563 or 15.63%
The variance is given by
Part A:
The expected return on an equally weighted portfolio of these three stocks is given by:
Part B:
Value of a portfolio invested 21 percent each in A and B and 58 percent in C is given by
For boom: 0.21(0.09) + 0.21(0.03) + 0.58(0.34) = 0.0189 + 0.0063 + 0.1972 = 0.2224 or 22.24%.
For bust: = 0.21(0.23) + 0.21(0.29) + 0.58(-0.14) = 0.0483 + 0.0609 - 0.0812 = 0.028 or 2.8%
Expected return = 0.66(0.2224) + 0.34(0.028) = 0.1468 + 0.00952 = 0.1563 or 15.63%
The variance is given by