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3 years ago

For every for female students in a training class there are two men what is the ratio of female to male students

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

5 0

For every for female students in a training class there are two men what is the ratio of female to male students.

Answer:

1:2

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Round off 3. 55 to one significant figure

Anika [276]

Answer:

3.6

Step-by-step explanation:

We must first clarify how a number is rounded.

To round a number to unity we have to look at the first number after the comma.

If this number is less than 5 (1, 2, 3, 4) we should not do anything, but if that number is 5 or greater (5, 6, 7, 8, 9) we must add a unit to the number.

That is to say:

<5 do nothing

=> 5 round to the next number (+1)

So in the case of 3.55 it would be.

3.55 = 3.6

8 0

3 years ago

Create an expression equivalent to 2 (3x =1) 3x+4 using the least number of terms

Mariana [72]

Answer:

$18x^2 + 18x -8$

Step-by-step explanation:

To use the least number of terms, simplify the expression. To simplify, distribute into the parenthesis and combine like terms.

$2(3x-1)(3x+4) = 2(9x^2-3x+12x-4) = 2(9x^2+9x-4) = 18x^2 + 18x -8$

6 0

3 years ago

The cheerleader section is on the basketball court. Each row requires a section of floor that is 1 ¾ yards by 3 ⅓ yards. How man

Lady_Fox [76]

Answer:

The area of the floor space is equal to 5.83 sq yards.

Step-by-step explanation:

Given that,

Each row requires a section of floor that is 1 ¾ yards by 3 ⅓ yards.

We need to find how many square yards of floor space are taken up by one row of cheerleaders.

We know that, the area of a rectangle is given by :

A = lb

So,

$A=\dfrac{7}{4}\times \dfrac{10}{3}\\\\A=5.83\ yards^2$

So, the area of the floor space is equal to 5.83 sq yards.

5 0

3 years ago

Consider the following points on cartesian plane. A (1;2) BC3; 1) and CG-3; 4) prove that Points A, B and C are collinear.

Arlecino [84]

Step-by-step explanation:

Collinear points have the same gradient

find the gradient(slope) between AB first

$= \frac{1 - 2}{3 - 1} \\ = \frac{ - 1}{2}$

Find the one for BC

$= \frac{4 - 1}{ - 3 - 3} \\ = \frac{3}{ - 6} \\ = \frac{ - 1}{2}$

Find the gradient between AC now

$= \frac{4 - 2}{ - 3 - 1} \\ = \frac{2}{ - 4} \\ = \frac{ - 1}{2}$

YOUR LAST STATEMENT NOW WILL BE

POINTS A,B AND C ARE COLLINEAR SINCE THE GRADIENTS BETWEEN THEM IS THE SAME!!!

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1 year ago

\left(x+3,\ y-6\right).

erastovalidia [21]

Say left is your mom and that would mean the answer would be 38922938829

3 0

2 years ago