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aivan3 [116]
3 years ago
7

there are 24 marbles in a bag: 2/16 of the marbles are red, 1/8 of the marbles are blue, 1/4 of the marbles are yellow, 1/3 of t

he marbles are black, 1/6 of the marbles are silver. how many of each type of marble is there?
Mathematics
1 answer:
Roman55 [17]3 years ago
4 0

Answer: 3 red marbles, 3 blue marbles, 6 yellow marbles, 8 black marbles, 4 silver marbles

Step-by-step explanation:

I know these are the correct number of marbles because I set up these numbers as fractions and solved for x.

For examples, 2/16 of the marbles are red. 2/16=x/24

You would do this because we are trying to find the number of marbles that are red out of the 24 marbles.

Next, we would do 24 divided by 16= 1.5 Then 1.5 times 2=3

You would follow this rule for each fraction of marbles in order to find the true number of marbles out of 24.

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Read each problem carefully and solve as required then answer the questions that follow
zepelin [54]

Answer:

Step-by-step explanation:

1) Number of children = c

Number of adults = a

Total ticket = 250

c + a = 250 -------------------(I)

Cost of 'c' children tickets = 200c

Cost of 'a' adult tickets =  450a

Total cost = 76000

200c + 450a = 76000  ------------------(II)

Multiply (I) by (-200) and then add

(I)*(-200)          -200c - 200a = -50000

(II)                    <u> 200c + 450a = 76000 </u>

<u> </u>                                    250a = 26000

                                           a = 26000/250

a = 104

Plug in a = 104 in equation (1)

c + 104 = 250

         c = 250 - 104

        c = 146

2)Number of children = 146

Number of adults = 104

7 0
3 years ago
If a truck can hold 4 boxes and if you needed to move 871 boxes across town , how many trips would you need to make?
Sauron [17]
218 trips is the amount 
5 0
3 years ago
Read 2 more answers
To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)
Pavlova-9 [17]

the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.


\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}


\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill

5 0
3 years ago
Show me how to solve -6+x/4= -5
schepotkina [342]
Multiply 4(-5) then add 6 and theres your X

7 0
3 years ago
Question 2<br> Solve by substitution.<br> 2x – 10y = 14<br> 2x + 6y = -2
Fed [463]

Answer:

x= 2 and y= -1

Step-by-step explanation:

x=7 +5y

14+16y=−2

y=−1

Substitute y=-1y=−1 into x=7+5yx=7+5y.

x=2

8 0
3 years ago
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