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Complete Question
An arts academy requires there to be 6 teachers for every 132 students and 4 tutors for every 44 students. How many students does the academy have per teacher? Per tutor? How many tutors does the academy need if it has 132 students?
Answer:
a)How many students does the academy have per teacher?
22 students per teacher
b) Per tutor?
11 students per tutor
c) How many tutors does the academy need if it has 132 students?
12 tutors
Step-by-step explanation:
a)How many students does the academy have per teacher?
6 teachers = 132 students
1 teacher = x
6 × x = 132 × 1
x = 132 × 1/6
x = 22 students
Hence, we have 22 students per tutor
b) Per tutor?
4 tutors = 44 students
1 tutor = x students
Cross Multiply
4 × x = 1 × 44
x = 44/4
x = 11 students
Hence, we have 11 students per tutor
c) How many tutors does the academy need if it has 132 students?
44 students = 4 tutors
132 students = x tutors
Cross Multiply
44 × x = 132 × 4
44x = 132 × 4
x = 132 × 4/44
x = 12 tutors
The academy needs 12 tutors if they have 132 students.
Answer:
Step-by-step explanation:
An = 54 + 3(n-1)
n = 1 ; A1 = 54 + 3*(1-1) = 54 + 3*0 = 54 +0 = 54
n =2 ; A2 = 54 + 3*(2-1) = 54 + 3*1 = 54 + 3 = 57
n =3 ; A3 = 54 + 3*(3-1) = 54 + 3*2 = 54 + 6 = 60
n =4 ; A4 = 54 + 3*(4-1) = 54 + 3*3 = 54 + 9 = 63
Answer:
x = -3
Step-by-step explanation:
to solve this equation you should first find the LCD or least common denominator for 2 and 6
the LCD is 6
so you should multiply each side by six
![6 * \frac{x + 5}{2} + 6 * -\frac{x + 3}{6} = 6 * 1](https://tex.z-dn.net/?f=6%20%2A%20%5Cfrac%7Bx%20%2B%205%7D%7B2%7D%20%2B%206%20%2A%20-%5Cfrac%7Bx%20%2B%203%7D%7B6%7D%20%3D%206%20%2A%201)
3x + 15 - x - 3 = 6
now combine like terms
2x + 12 = 6
now subtract both sides by 12
2x = -6
lastly divide everything by 2
x = -3
that is your answer
The common difference of the data set of A is 1. While the common difference in the data set of B is 0.80
B.) 3.8 + 0.80 = 4.60
4.6 + 0.80 = 5.40
5.4 + 0.80 = 6.20
6.2 + 0.80 = 7.00