7 students brought lunch from home
Answer:
2 solutions
Step-by-step explanation:
I like to use a graphing calculator to find solutions for equations like these. The two solutions are ...
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To solve this algebraically, it is convenient to subtract 2x-7 from both sides of the equation:
3x(x -4) +5 -x -(2x -7) = 0
3x^2 -12x +5 -x -2x +7 = 0 . . . . . eliminate parentheses
3x^2 -15x +12 = 0 . . . . . . . . . . . . collect terms
3(x -1)(x -4) = 0 . . . . . . . . . . . . . . . factor
The values of x that make these factors zero are x=1 and x=4. These are the solutions to the equation. There are two solutions.
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<em>Alternate method</em>
Once you get to the quadratic form, you can find the number of solutions without actually finding the solutions. The discriminant is ...
d = b^2 -4ac . . . . where a, b, c are the coefficients in the form ax^2+bx+c
d = (-15)^2 -4(3)(12) = 225 -144 = 81
This positive value means the equation has 2 real solutions.
Answer:
-10
Step-by-step explanation:
2*-2=-4
3*-5=-15
1-(-4)+(-15)=-10
Answer:
A box plot is drawn with end points at 24 and 49.The box extends from 28 to 44 and a vertical line is drawn inside the box at 34.
Step-by-step explanation:
Ordering the data given :
24,28,32,34,40,44,49
We can calculate the 5 number summary required to give the appropriate boxplot that can be produced :
Minimum = 24
Maximum = 49
Median = 1/2(n+1)th term
n = 7
Median = 1/2(8) = 4th term
Median = 34
Lower quartile, Q1 = 1/4(n+1)th term
n = 7
1/4(8) = 2nd term
Q1 = 28
Upper quartile : 3/4(n+1)th term
n = 7
Q3 = 3/4(8) = 6th term
Q3= 44