Ask question

Ask question

All categories

3 years ago

5

An investor has an account with stock from two different companies. Last year, his stock in Company A was worth $7200 and his st

ock in Company B was worth $3510. The stock in Company A has decreased 24% since last year and the stock in Company B has decreased 20%. What was the total percentage decrease in the investor's stock account? Round your answer to the nearest tenth (if necessary)

1 answer:

djyliett [7]3 years ago

7 0

Answer:

22.7%

Explanation:

Given that last year stock for company A was $7200

The stock for company B last year was worth $3510

Stock in company A decreased by 24%

This means the new value of stock for company A became;

(100-24)/100 *$7200

76/100*$7200

0.76*7200 =$5472

Stock in company B decreased by 20%

This means the new value of stock for company B became;

(100-20)/100 *$3510

80/100*$3510

0.8*$3510

$2808

Original investors stock value was = $7200+$3510 =$10710

New investors stock value is = $5472+$2808=$8280

Decrease in value of stock = $10710-$8280 =$2430

percentage decrease in stock value = decrease in stock/original value of stock *100%

=2430/10710 *100 =22.689

=22.7%

You might be interested in

What is an inference?

sp2606 [1]

Answer:

A conclusion derived from evidence and logical reasoning

Explanation:

5 0

3 years ago

Read 2 more answers

A baker is decorating cupcakes using 3 cherries for every 1 cupcake. If the baker has 12 cherries and 5 cupcakes, what is the th

11111nata11111 [884]

C. 4 decorated cupcakes.

Because 12 divided by 3 = 4, so you can have 4 cupcakes each with 3 cherries

7 0

3 years ago

Read 2 more answers

A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55

Xelga [282]

<u>Answer:</u> The volume of NaOH required is 402.9 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For HCl:</u>

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol$

<u>For sulfuric acid:</u>

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

$0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol$

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

$0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL$

Hence, the volume of NaOH required is 402.9 mL

6 0

3 years ago

It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir wi

Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d = $4.50\times 10^1 m=45 m$

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

$V=\pi r^2 h$

$=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L$

$1 m^3=1000 l$

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

$m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg$

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

5 0

3 years ago

sineoko [7]

Answer:

this answer is b no u can it very easy

Explanation:

because solid and liquid can compress easily

plz support me

4 0

3 years ago

Read 2 more answers