Fiona wrote the linear equation y = y equals startfraction 2 over 5 endfraction x minus 5.x – 5. when henry wrote his equation, they discovered that his equation had all the same solutions as fiona’s. which equation could be henry’s? x – x minus startfraction 5 over 4 endfraction y equals startfraction 25 over 4 endfraction. y = x – x minus startfraction 5 over 2 endfraction y equals startfraction 25 over 4 endfraction. y = x – x minus startfraction 5 over 4 endfraction y equals startfraction 25 over 2 endfraction. y = x – x minus startfraction 5 over 2 endfraction y equals startfraction 25 over 2 endfraction. y =

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5

2 equivalent expressions to represent the area of a rectangle and the top is n-3 and the side is 2

area of the rectangle = length * width

given length = n - 3

width = 2

replace it in the area

so area of the rectangle = (n-3) *2

= 2(n-3)

one of the expression is 2(n-3)

to get another expression we distribute 2 inside the parenthesis

2(n-3) is 2n - 6

so another expression is 2n - 6

two equivalent expression are

2(n - 3)   , 2n - 6