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3 years ago

The ratio of boys to girls in a class is 4:5 what fraction of the class is boys and what fraction of the class is girls

2 answers:

7 0

First, let's start off with the information we already have. Since the ratio is 4 boys to 5 girls, there has to be a minimum of 9 students (4 being boys and 5 being girls).

9 will be the denominator of both of our fractions since the boys and girls are in the same class.

The fraction of boys in the class is 4/9 (since there are 4 boys to 5 girls in a class of 9, we would write 4 over 9) and the fraction of girls in the class is 5/9.

This is a simplified version of the fractions. If this is a multiple-answer question and 4/9 and 5/9 are not up there, try multiplying the fractions with different numbers and see what fractions will be correct.

But either way, 4/9 and 5/9 are correct :)

LenKa [72]3 years ago

4 0

Answer:

4/9 and 5/9

Step-by-step explanation:

It's simple really you add up the ratios 4+5=9

So 9 will be the amount you have so this will be the denominator.

This means the fraction of boys will be 4/9 and the fraction of girls will be 5/9.

I hope this helped :)

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5 0

3 years ago

If the diver in item 2 leaves the board with an initial upward speed of 2.00 m/s, find the diver's speed when striking the water

Taya2010 [7]

Answer:

$v_f=9.0774m/s$

Step-by-step explanation:

Let's assume

initial velocity =vi

final velocity=vf

initial height =hi

final height =hf

acceleration due to gravity =g

we can use formula

$\frac{1}{2}v_i^2+gh_i= \frac{1}{2}v_f^2+gh_f$

we are given

$v_i=2m/s$

$h_i=2\times 2=4$

$h_f=0$

$g=9.8m/s^2$

now, we can plug values

$\frac{1}{2}(2)^2+(9.8)(4)= \frac{1}{2}v_f^2+(9.8)(0)$

now, we can solve for vf

$\frac{1}{2}(2)^2+(9.8)(4)= \frac{1}{2}v_f^2$

we get

$v_f=9.0774m/s$

4 0

3 years ago

Hello, can someone help me with this problem?

exis [7]

Answer:

Area of Rectangle A

$Area = 4x^2$

Area of Rectangle B

$Area = 2x^2$

Fraction

$Fraction =\frac{2}{3}$

Step-by-step explanation:

From the attached, we understand that:

The dimension of rectangle A is 2x by 2x

The dimension of rectangle B is x by 2x

Area of rectangle is calculated as thus;

$Area = Length * Breadth$

Area of Rectangle A

$Area = 2x * 2x$

$Area = 4x^2$

Area of Rectangle B

$Area = x * 2x$

$Area = 2x^2$

Area of Big Rectangle

The largest rectangle is formed by merging the two rectangles together;

The dimension are 3x by 2x

The Area is as follows

$Area = 2x * 3x$

$Area = 6x^2$

The fraction of rectangle A in relation to the largest rectangle is calculated by dividing area of rectangle A by area of the largest rectangle;

$Fraction = \frac{Rectangle\ A}{Biggest}$

$Fraction =\frac{4x^2}{6x^2}$

Simplify

$Fraction =\frac{2x^2 * 2}{2x^2 * 3}$

$Fraction =\frac{2}{3}$

7 0

3 years ago

Which of the following is true?

gtnhenbr [62]

Answer:

C..... I think it is a answer

6 0

3 years ago

Z = { x:x is an integer, x ≥ - 3 and x ≤ + 3}

tankabanditka [31]

Answer:

If this is asking for the set:

Z = {-3, -2, -1, 0, 1, 2, 3}

Step-by-step explanation:

Z is the set of all integers and it appears that you are being asked for the values in the set Z that are within the range: -3 ≤x ≤ 3

5 0

2 years ago