Quadratic formula is x=(-b+-sqrt(b^2-4ac))/(2a)
a= 1, b=-4, and c=3
Therefore, x=(-(-4)+-sqrt((-4)^2-4(1)(3))/(2(1))
Solve for both the plus and minus equations; the answers will be the roots to the quadratic equation.
Answer:
5
Step-by-step explanation:
The interquartile range is the difference between the upper quartile and the lower quartile.
First find the median.
The median is the middle value of the data set arranged in ascending order
1 5 5 7 9 ← data in ascending order
↑ median
The lower quartile is the middle value of the data to the left of the median. If there is not an exact middle then it the average of the values on either side of the middle.
1 5
↑ lower quartile =
= 3
The upper quartile is the middle value of the data to the right of the median.
7 9
↑ upper quartile =
= 8
Thus
interquartile range = 8 - 3 = 5
The error is that 9 shouldnt be written as nine it should be written as 3x3 or 3^2. So the answer should be 2^3 x 3^2 because 9 can be divided into even more prime factors of 72.
Step-by-step explanation:
y = mx + b
y = -7x + 7
this assumes m = - 7 , it got chopped off, otherwise what ever slope is change the first 7 for that number.