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Levart [38]
2 years ago
10

A simple index of three stocks have opening values on day 1 and day 8 as shown in the table below. What is the rate of change of

this simple index over one week? Round your awnser to the nearest tenth.

Mathematics
2 answers:
Neporo4naja [7]2 years ago
6 0

This isn't a stock index as much as it is a portfolio, but it doesn't matter.

On day one the assets are worth

8000*4.25+5000*2.90+2000*6.40 = $61,300

On day 8

8000*3.90 + 5000*2.50 + 2000*6.10 = $55,900

Change: 55900/61300 - 1 = -.08809

That seems like none of the above but we'll choose

choice B

Marianna [84]2 years ago
3 0

Answer:

–7.7%

Step-by-step explanation:

On apex!!

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A rectangular portrait measures 50cm by 70cm. It is surrounded by a rectangular frame of uniform width. If the area of the frame
snow_tiger [21]

Let us assume uniform width = x cm wide.

Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.

Therefore,  length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be =  (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame -  area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

 3500 = (2x+50) * (2x+70) - 3500.

Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

4x^2 +240x -3500 = 0.

Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula =\frac{-b\pm \sqrt{b^2-4ac}}{2a}, we get

=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}

x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)

x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)

We cant take negative value.

So, x=5\left(\sqrt{71}-6\right)=12.13

We could take approximately 12 cm.



7 0
2 years ago
Bonzo went to a carnival. At the first
Shalnov [3]

Bonzo started with 2.1 dollars

Step-by-step explanation:

Assume that Bonzo has $x

1. Change $x to cents

2. Calculate the remaining money with him after each game

3. Equate the left money in the 3rd game by zero to find x

∵ $1 = 100 cents

∴ $x = 100 x cents

First game

∵ Bonzo has 100 x cents

∵ He paid 10¢ to get in

∴ The money left is (100 x - 10)

∵ He spent half  the money he had left

- That mean the money left with him is the other have

∴ The money left with him = \frac{1}{2} (100 x - 10) = (50 x - 5) cents

∵ He spent 10¢  to get out

∴ The money left after 1st game = (50 x - 5) - 10

∴ The money left after 1st game = (50 x - 15) cents

Second game

∵ Bonzo has (50 x - 15) cents

∵ He paid 10¢ to get in

∴ The money left is (50 x - 15) - 10 = (50 x - 25)

∵ He spent half  the money he had left

∴ The money left with him = \frac{1}{2} (50 x - 25) = (25 x - 12.5) cents

∵ He spent 10¢  to get out

∴ The money left after 2nd game = (25 x - 12.5) - 10

∴ The money left after 2nd game = (25 x - 22.5) cents

Third game

∵ Bonzo has (25 x - 22.5) cents

∵ He paid 10¢ to get in

∴ The money left is (25 x - 22.5) - 10 = (25 x - 32.5)

∵ He spent half  the money he had left

∴ The money left with him = \frac{1}{2} (25 x - 32.5) = (12.5 x - 16.25) cents

∵ He spent 10¢  to get out

∴ The money left after 3rd game = (12.5 x - 16.25) - 10

∴ The money left after 3rd game = (12.5 x - 26.25) cents

∵ He found he had no money left after the 3rd game

∴ Equate the left money with him by zero

∴ 12.5 x - 26.25 = 0

- Add 26.25 to both sides

∴ 12.5 x = 26.25

- Divide both sides by 12.5

∴ x = 2.1

<em>Bonzo started with 2.1 dollars </em>

Learn more:

You can learn more about money in  brainly.com/question/1870710

#LearnwithBrainly

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