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jek_recluse [69]
3 years ago
8

ASAP I NEED HELP

Mathematics
1 answer:
alexdok [17]3 years ago
7 0

The vertex of this parabola is at (3,-2). When the x-value is 4, the y-value is 3: (4,3) is a point on the parabola. Let's use the standard equation of a parabola in vertex form:

y-k = a(x-h)^2, where (h,k) is the vertex (here (3,-2)) and (x,y): (4,3) is another point on the parabola. Since (3,-2) is the lowest point of the parabola, and (4,3) is thus higher up, we know that the parabola opens up.

Substituting the given info into the equation y-k = a(x-h)^2, we get:

3-[-2] = a(4-3)^2, or 5 = a(1)^2. Thus, a = 5, and the equation of the parabola is

y+2 = 5(x-3)^2 The coefficient of the x^2 term is thus 5.

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Interquartile range (IQR) = third quartile (Q3) - first quartile (Q1)

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Q3 = the middle value of the upper part of the data set, from the median to your right.

The median lies between the 5th and 6th value that is enclosed in the parenthesis below:

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IQR = $22 - $15 = $7

Range = highest amount - least amount = 25 - 12 = $13

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