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ra1l [238]
1 year ago
9

tenemos $5000 en una cuenta. A final de cada mes se ingresa un 5% del dinero que hay en la cuenta en dicho momento. Calcular el

dinero que habrá en la cuenta después de un trimestre¿ en que porcentaje ha subido la cantidad inicial?
Mathematics
1 answer:
LiRa [457]1 year ago
6 0

Al final del primer mes tenemos:

5000 + 5000*0.05 = 5250

Al final del segundo mes:

5250 + 5250*0.05 = 5512.5

Y al final del tercer mes nos queda:

5512.5 + 5512.5*0.05 = 5788.125

El dinero que habra en la cuent despues de un trimestre es $5788.125

Para calcular en que porcentaje ha subido la cantidad inicial, dividimos la cantidad extra obtenida despues de un trimestre entre la cantidad inicial:

\begin{gathered} \frac{788.125}{5000}=0.157625 \\ \text{Expresado en porcentaje:} \\ 0.157625\cdot100=15.7625 \end{gathered}

Ha subido en un 15.7625%

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Answer:

30 students

Step-by-step explanation:

80%  of  N   =   24            where N  is  the total number of students

So

.80 N   =  24

N =  24 /  .80     =      30

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6 0
3 years ago
Marta made a fruit salad with 30 strawberries and 50 grapes. Assuming each strawberry contains 6 calories and each grape contain
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Okay so i am no math genius but i did it like this, 30 strawberries and 50 grapes would be 80 fruits. say all 5 people got an equal amount so 80/5=16. so 16 pieces would be 8 each so 8 strawberries(sb) and 8 grapes(g). sb= 6 calories and g=3 calories. 48 calories for sb and 24 for g. 72 calories each. unless i did this totally wrong
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3 years ago
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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
a rectangular computer screen has an area o A square inches. The width of the computer is 7 inches. Which equation rpresents x t
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Answer:

x = A/7

Step-by-step explanation:

Given that the screen is rectangular, recall that the area A of a rectangle whose length is L and width is W is given as

A = L * W

Hence if the width is 7 inches and the length is x inches then the area which is A square inches is connected to both sides as

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Divide both sides by 7

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A rectangular painting's width is 6 less than twice its length. Given that the perimeter of the painting is 84 inches, select al
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Answer:

Step-by-step explanation: it is

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