A)
Let x represent the cost of 1 student, and y the cost of 1 teacher.
B)
In the first group, there's 25 students and 2 teachers. Their total cost is $97.50
So 25x + 2y = 97.50
In the second group, there's 32 students and 3 teachers. Their total cost is $127
So 32x + 3y = 127
We get the following system of equations:
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
C)
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
In equation (1)
25x + 2y = 97.50
25x + 2y - 2y = 97.50 - 2y
25x = 97.50 - 2y
25x / 25 = 97.50/25 - 2y/25
x = 3.9 - (2/25)y
In equation (2), let's replace x by its algebraic value
32x + 3y = 127
32(-2/25y + 3.9) + 3y = 127
11/25y + 124.8 = 127
11/25y + 124.8 - 124.8 = 127 - 124.8
11/25y = 2.2
(11/25y) / (11/25) = 2.2 / (11/25)
y = 5
x = -2/25y + 3.9
x = -2/25 * 5 + 3.9
x = 3.5
So the cost of each student is $3.5, and the cost of each teacher is $5.
Hope this helps! :)
Answer:
0.91517
Step-by-step explanation:
Given that SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certificate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random.
Let A - the event passing in SAT with atleast 1500
B - getting award i.e getting atleast 1350
Required probability = P(B/A)
= P(X>1500)/P(X>1350)
X is N (1100, 200)
Corresponding Z score = 

Since these angles clearly form a circle and a circle is 360 degrees,
120-x+2x+16+3x-72=64+4x=360. Subtracting 64 from both sides, we get
4x=296 and by dividing both sides by 4 we get x=74
QUESTION 1
The point C(3.6, -0.4) divides AB in the ratio 3 : 2.
The coordinates of A are (-6, 5).
Let the coordinates of B be 
We use the formula:
to determine the x-coordinate of B.
We substitute the known values to obtain:





This implies that:

We also use the formula:
to find the y-coordinate.






The coordinates of B are (10,-4)
QUESTION 2.
If point D divides CD in the ratio 4 : 5.
Then the coordinates of D are:


The coordinates of D are 