Answer:
She should order a total of 5,160 medium sized t-shirts
Step-by-step explanation:
To calculate the number of medium sized t-shirt to be ordered, we shall be using the proportion of students that wanted medium sized t-shirt in her survey to multiply the total number of students that we have.
From the question, we can identify that the probability that a student will like a medium sized t-shirt is simply 129/300
Now we have about 12000 students in the University, using the probability from the survey, the number of medium sized t shirt she should order would be;
129/300 * 12,000 = 129 * 40 = 5,160 medium sized t-shirts
Answer:
70 + 4D
Step-by-step explanation:
- (2 x 100) + (7 x 10) + (4 x D) + (2 x 100)
- (200) + (70) + 4D + 200
-200 + 70 +4D + 200
-130 + 4D + 200
70 + 4D
Answer:
-2x-10
Step-by-step explanation:
just distribute the -2 :)
Answer:
Step-by-step explanation:
<u>Similar Triangles</u>
By looking at the construction of the figure, we can safely assume both triangles are similar, i.e. their internal angles are equal and their sides are proportional. Following the proportion of the heights and bases of both triangles we can set this relationship:
Simplifying both fractions
Solving for x
Answer:
Step-by-step explanation:
A system of linear equations is one which may be written in the form
a11x1 + a12x2 + · · · + a1nxn = b1 (1)
a21x1 + a22x2 + · · · + a2nxn = b2 (2)
.
am1x1 + am2x2 + · · · + amnxn = bm (m)
Here, all of the coefficients aij and all of the right hand sides bi are assumed to be known constants. All of the
xi
’s are assumed to be unknowns, that we are to solve for. Note that every left hand side is a sum of terms of
the form constant × x
Solving Linear Systems of Equations
We now introduce, by way of several examples, the systematic procedure for solving systems of linear
equations.
Here is a system of three equations in three unknowns.
x1+ x2 + x3 = 4 (1)
x1+ 2x2 + 3x3 = 9 (2)
2x1+ 3x2 + x3 = 7 (3)
We can reduce the system down to two equations in two unknowns by using the first equation to solve for x1
in terms of x2 and x3
x1 = 4 − x2 − x3 (1’)
1
and substituting this solution into the remaining two equations
(2) (4 − x2 − x3) + 2x2+3x3 = 9 =⇒ x2+2x3 = 5
(3) 2(4 − x2 − x3) + 3x2+ x3 = 7 =⇒ x2− x3 = −1
