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alukav5142 [94]
2 years ago
12

A bag contains 8 red marbles, 9 yellow marbles, and 7 green marbles. how many additional red marbles must be added to the 24 mar

bles already in the bag so that the probability of randomly drawing a red marble is 3/5
Mathematics
2 answers:
Mice21 [21]2 years ago
8 0

Answer:

16 red marbles

Step-by-step explanation:

8 red

9 yellow

7 green

total = 8 + 9 + 7

p(red) = 8/24 = 1/3

Add r number of red marbles to the total.

8 + r red

9 yellow

7 green

total = 8 + r + 9 + 7 = r + 24

p(red) = (8 + r)/(r + 24) = 3/5

Cross multiply.

40 + 5r = 72 + 3r

2r = 32

r = 16

Answer: 16 red marbles

EleoNora [17]2 years ago
6 0

Answer:

16

Step-by-step explanation:

bag red: 8 + x

yellow:9

green 7/24

3/5 = 8+ x/24+ x

5(8 + x) = 3 (24 + x)

40 + 5x

-40 -3x = 72 + 3x

-40 -3x    -40 - 30

2x/2 = 32/2  so,x = 16

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3 years ago
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Answer:

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Step-by-step explanation:

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F(x) = 2/x, g(x) = |x| -1 <br><br><br> What is the domain of (fog) (x)?<br><br> will mark brainliest
brilliants [131]

Answer:

x≠-1,1

Step-by-step explanation:

f(g(x)) is a composition where g(x) is is substituted for x in f(x).

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3 0
3 years ago
the perimeter of a rectangle is 55 inches. the ratio of the length to width is 7:4. use proportions to find the area of the rect
Leni [432]

Answer:

Step-by-step explanation:

let : x the length and y the width    .... ( x  > y and x ; y reals numbers )

the perimeter is : 2(x+y)

2(x+y) = 55...(1)

x/y = 7/4 ...(2)

by (2) : x = (7/4)y

subsct in (1) : 2( (7/4)y +y ) =55

(7/2)y +2y =55

multiply by 2

7y +4y = 110

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4 0
3 years ago
A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).
maksim [4K]

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill

B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}

\stackrel{\textit{\large perimeter of ABC}}{4\sqrt{2}+\sqrt{13}+\sqrt{53}~~\approx~~ 16.54}

3 0
2 years ago
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