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stira [4]
3 years ago
9

You have 2 positive numbers. One number is one-fifth of the other number. The difference between the two numbers is 228, find th

e numbers.
Mathematics
1 answer:
liberstina [14]3 years ago
7 0

Hey there! I'm happy to help!

Let's call these numbers x and y and model the information we have.

x=1/5y

y-x=228 (we see that x is smaller than y in the first equation, so we put it second )

We see that x is equal to 1/5y. We can switch the x in the second equation with 1/5y and solve for y.

y-1/5y=228

4/5y=228

We divide both sides by 4/5.

y=285

We know that x is 1/5 of y, so we divide by 5.

285/5=57

x=57

So, our numbers are 57 and 285.

Have a wonderful day! :D

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45% of 450 is 202.50

$202.50 is the price of markup

$652.50 is the selling price

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Two standard number cubes are thrown. Find the probability that the sum of the numbers showing is ten given that the first numbe
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I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Raymond would like to go to rebounders trampoline center. The total cost , c, for every hour spent boucing, h, is representing b
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Answer:

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Para reunir dinero para su gira de estudios , los alumnos de un curso deciden vender números de una rifa que se encuentran numer
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Respuesta:

0.53

Explicación:

Para calcular la posibilidad del evento A: "ganar la rifa comprando todos los números múltiplos de 3 o 5", debemos usar la siguiente fórmula.

P(A) = casos favorables / casos posibles

Evaluemos primero todos los casos que son múltiplos de 3, entre 1 y 100: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99. En total son 33.

Ahora, evaluemos todos los casos que son múltiplos de 3, entre 1 y 100: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100. En total son 20.

El número total de casos favorables es 33 + 20 = 53.

El número de casos posibles es el total de números de 1 a 100, es decir 100.

Luego P(A) = 53/100 = 0.53.

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Both expressions are = to 11 when x is 2.

Therefore, the expressions are equivalent.

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