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Mekhanik [1.2K]
3 years ago
7

On its first day of trading, Twitter closed at $41.57 per share. Two years later and the price was $26.85, what was the annual r

eturn on the stock if returns are compounded daily?
A. 21.86%.
B. -21.85%.
C. -5.99%.
D. -19.63%.
Business
1 answer:
Mkey [24]3 years ago
8 0

Answer:

B. -21.85%.

Explanation:

Calculation for the annual return on the stock

First step is to calculate the Number of periods

Number of periods = 2 * 365 days in a year

Number of periods= 730

Second Step is to calculate the Daily return using this formula

Daily return = (Future value / initial value)^1/n - 1

Let plug in the formula

Daily return = (26.85 / 41.57)^1/730 - 1

Daily return = (0.645898)^1/730 - 1

Daily return = 0.999401 - 1

Daily return = -0.00059861*100

Daily return = -0.059861%

Last step is to calculate annual return

Using this formula

Annual return=Daily return/ Numbers of days in a year

Annual return = -0.059861% * 365

Annual return = -21.85%

Therefore the annual return on the stock if returns are compounded daily will be 21.85%

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C

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Vornoda Inc., a multinational clothing and accessory brand, has been facing huge economic losses due to unpredictable exchange r
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Answer:

consider opening manufacturing companies in each nation

Explanation:

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I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

8 0
2 years ago
A metal sphere of radius 15 cm has a net charge of 3.0 $ 10#8 C. (a) What is the electric field at the sphere’s surface? (b) If
denis-greek [22]

Answer:

E = 1.20 × 10^{4} N/C

V = 1800 V

x = 0.058 m

Explanation:

given data

radius = 15 cm

net charge = 3 × 10^{-8} C

electric potential decreased = 500 V

solution

we get here electric field at the sphere’s surface that is

electric field at the sphere’s surface E  =  \frac{q}{4\pi \epsilon _o R^2}   ............1

put here value

electric field at the sphere’s surface E  = \frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15^2}    

E = 1.20 × 10^{4} N/C

and

potential on surface of sphere is

V =  \frac{q}{4\pi \epsilon _o R} ................2

V = \frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15}  

V = 1800 V

and

now we get distance that is x

and we know here

ΔV = V(x) - V   ..............3

substitute here value

-500V = \frac{q}{4\pi \epsilon _o } \times (\frac{1}{R+x} - \frac{1}{R})

-500 V = {3\times 10^{-8}\times 8.99\times 10^9} \times (\frac{1}{0.15+x} - \frac{1}{0.15})

solve it we get x

x = 0.058 m

3 0
3 years ago
Beckett Co. sold merchandise on account for $11,700, terms 1/10, n/30. Freight charges of $360 were prepaid by the seller and ad
Yuki888 [10]

Answer:

Beckett Co received $   10.699,65‬

Explanation:

We have to determnate the balance of the invoice after the returned and the discount granted. We also have to receive the freight from the customer as we prepaid them to refund this amount against the customer:

invoice balance:

11,700 nominal less 1,350 return = 10,350

discount granted: 10,350 x 1% =           (10.35)

freight refund                               <u>       360        </u>

total cash proceeds:                      10.699,65‬

3 0
3 years ago
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