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3 years ago

The ratio of students to teachers at a school is 19 to 1 how many students are there in each total number of people

Mathematics

1 answer:

vladimir2022 [97]3 years ago

5 0

Question:

The ratio of students to teachers at a school is 19 : 1. How many students are there in each total number of people?

A: 100 people

b. 280 people

c.440 people

d.760 people

Answer:

A )95 students and 5 teachers

B)266 students and 14 teachers

C)418 students and 22 teachers

D)722 students and 38 teachers

Step-by-step explanation:

A: 100 people

Ratio = 19 : 1

The sum of the ratio is 19 + 1 = 20

Scale factor = $\frac{100}{20}$

scale factor = 5

The number students is $19 \times 5$ = 95

The number of teachers is $1 \times 5$ = 5

b. 280 people

Ratio = 19 : 1

The sum of the ratio is 19 + 1 = 20

Scale factor = $\frac{280}{20}$

scale factor = 14

The number students is $19 \times 14$ = 266

The number of teachers is $1 \times 14$ = 14

c.440 people

Ratio = 19 : 1

The sum of the ratio is 19 + 1 = 20

Scale factor = $\frac{440}{20}$

scale factor = 22

The number students is $19 \times 22$ = 418

The number of teachers is $1 \times 22$ = 22

d.760 people

Ratio = 19 : 1

The sum of the ratio is 19 + 1 = 20

Scale factor = $\frac{760}{20}$

scale factor = 38

The number students is $19 \times 38$ = 722

The number of teachers is $1 \times 38$ = 38

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3 years ago

State the equation of the line that is perpendicular to 4x-3y=10 through the point (-2,4)

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The equation of the line that is perpendicular to 4x - 3y = 10 through the point (-2,4) is $y-4=\frac{-3}{4}(x+2)$

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Given, line equation is 4x – 3y = 10

We have to find a line that is perpendicular to 4x – 3y = 10 and passing through (-2, 4)

Now, let us find the slope of the given line,

$\text { Slope of a line }=\frac{-\mathrm{x} \text { coefficient }}{\mathrm{y} \text { coefficient }}=\frac{-4}{-3}=\frac{4}{3}$

We know that, slope of a line $\times$ slope of perpendicular line = -1

$\begin{array}{l}{\text { Then, } \frac{4}{3} \times \text { slope of perpendicular line }=-1} \\\\ {\rightarrow \text { slope of perpendicular line }=-1 \times \frac{3}{4}=-\frac{3}{4}}\end{array}$

Now, slope of our required line = $\frac{-3}{4}$ and it passes through (-2, 4)

The point slope form is given as:

$\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }} \\\\ {\text { Here in our problem, } m=-\frac{3}{4}, \text { and }\left(x_{1}, y_{1}\right)=(-2,4)} \\\\ {\text { Then, line equation } \rightarrow y-4=-\frac{3}{4}(x-(-2))}\end{array}$

$y-4=\frac{-3}{4}(x+2)$

Hence the equation of line is found out

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