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3 years ago

A fraction has a denominator that is 4 greater than its numerator. It is equivalent to 9/15. What is the fraction?

2 answers:

7 0

Et the numerator be x, so the denominator is x + 4. So we have:

x/(x + 4) = 9/15 (cross multiply) 
15x = 9(x + 4) 
15x = 9x + 36 
15x - 9x = 9x - 9x + 36 
6x = 36 
6x/6 = 36/6
x = 6
x + 4 = 6 + 4 = 10

Sine the numerator is 6 and the denominator is 10, then the fraction is 6/10, which in the simplest form is equal to 3/5. 

Since 9/15 in the simplest form is also equal to 3/5, then 6/10 is equivalent to 9/15. 

Thus, 6/10 is the required fraction. hope i helped

Rasek [7]3 years ago

6 0

$\frac{x}{x + 4} = \frac{9}{15}$
$\frac{x}{x + 4} = \frac{3}{5}$
$5x = 3(x + 4)$
$5x = 3(x) + 3(4)$
$5x = 3x + 12$
$5x - 3x = 3x - 3x + 12$
$\frac{2x}{2} = \frac{12}{2}$
$x = 6$

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Your team wins 18 medals at a competition. The metals are gold, silver, and bronze in a ratio of 2:4:3. How many of each metal d

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Answer

Find out the how many of each metal did your team win.

To prove

As given

Your team wins 18 medals at a competition.

The metals are gold, silver, and bronze in a ratio of 2:4:3.

Let us assume that x be the scalar multiple of the gold, silver and bronze medal.

Thus

Number of gold medal = 2x

Number of silver medal = 4x

Number of bronze medal = 3x

Than the equation becomes

2x + 4x + 3x = 18

9x = 18

$x = \frac{18}{9}$

x = 2

Number of gold medal = 2 × 2

= 4

Number of silver medal = 4 × 2

= 8

Number of bronze medal = 3 × 2

= 6

Therefore the number of gold medal are 4, silver medal are 8, bronze medal are 6.

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3 years ago

At the school carnival the sixth graders are making directional arrows each arrow is to be painted red how much area needs to be

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We can divide arrow in two sections:

A 1 ( rectangle ) = 4 x 5 = 20 sq in.

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A = A 1 + A 2 = 20 + 9 = 29 sq in

Answer: B )29 sq in.

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3 years ago

Read 2 more answers

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

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Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

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