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hram777 [196]
4 years ago
11

you save $5,000.00 and invest 60% of it in stocks while leaving the rest in a savings account earning a 4.9% APR. The stock incr

eases 9% in the first year and loses 4% of its value the second year. What is the total amount gained during the 2 years? What would the gain have been if all of the investment had been left in the savings account?
Mathematics
1 answer:
Nataly_w [17]4 years ago
5 0
Amount invested in stocks
5,000×0.60
=3,000
after one year gains 9%
3,000×(1+0.09)
=3,270
After second year loses 4%
3,270×(1−0.04)
=3,139.2 amount after second year
So Stocks gained 139.2 (3139.2-3000)

Amount of saving account
5,000×0.40
=2,000
After 2 years
2,000×(1+0.049)^(2)
=2,200.802
So it gained 200.802 (2200.802-2000)

Total amount after 2 years
3,139.2+2,200.802
=5,340.002
Gained 340.002 (5340.002-5000)
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yaroslaw [1]

Answer:

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3 0
3 years ago
10•1/2+(-6)(-3) Simplified in steps
dangina [55]

Answer:

23

Step-by-step explanation:

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3 years ago
Help me with this....After drawing a histogram....How do u find the class in which the median lies in a histogram....​
Ann [662]

Answer:

Get all the values of the histogram and put them in order:

For example 2,5,7,1,7,3  becomes 1,2,3,5,7,7 and then you get the middle number, which is in this case (3+5)/2=3.5

Then see which of the readings lie on 3.5.

Step-by-step explanation:

5 0
3 years ago
solve for x and y ,32base x+51base y and 23base x +42 base y =710. ,number bases interms of simultaneous equation​
In-s [12.5K]

Answer:

<h2>x = -348.5, y = 350.5</h2>

Step-by-step explanation:

32base x means 3x + 2

Similarly, 32base x+51base y equals to 3x + 2 + 5y + 1 = 3x + 5y + 3

and 23base x +42 base y means 2x + 3 + 4y + 2 = 2x + 4y + 5

As per the given condition, 3x + 5y + 3 = 2x + 4y + 5\\x + y = 2

Putting, y = 2 - x in 2x + 4y + 5 = 710, we get

2x + 4y = 705\\2x + 4(2 - x) = 705\\8 - 2x = 705\\4 - x = 352.5\\x = -348.5

y = 2 - x = 2 - (-348.5) = 2 + 348.5 = 350.5

6 0
3 years ago
Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

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You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0
2 years ago
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