All categories

3 years ago

A college has a student to teacher ratio of 35 to 2. If there are 170 teachers at the college, how many students attend the

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

7 0

Answer: 2975

Step-by-step explanation:

From the question, we are informed that a college has a student to teacher ratio of 35 to 2 and that there are 170 teachers at the college,

The number of students in the college will then bee calculated as:

= 35 × 170/2

= 35 × 85

= 2975

Therefore, 2975 students attend the college

You might be interested in

Evaluate 0.035×1.02/0.00015 leaving your answer in standard form

denis-greek [22]

0.035 x 1.02 is equal to 0.0357. 0.0357 divided by 0.00015 is equal to 238.

3 0

3 years ago

Pls help, question on the picture

adelina 88 [10]

What is the question

7 0

3 years ago

At the gym, Jasper was able to bench press 224 pounds, which was StartFraction 7 over 8 EndFraction of the weight that Balin was

myrzilka [38]

Answer:

Balin could bench press 256 pounds.

Step-by-step explanation:

Jasper was able to bench press 224 pounds, which is 7/8 of x, which is the weight that Balin was able to bench press. So, mathematically:

$\frac{7}{8}x = 224$

$7x = 224*8$

$x = \frac{224*8}{7} = 256$

Balin could bench press 256 pounds.

4 0

3 years ago

Read 2 more answers

Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2

fredd [130]

Answer:

Both get the same results that is,

$\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

Step-by-step explanation:

Given :

$\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

and initial population,

$\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]$

a) - After two times, we will find in each position.

$P_2=[P].[M]^2=[P].[M].[M]$

$M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

$=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]$