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BartSMP [9]
2 years ago
13

Problem 5-1 Simple Interest versus Compound Interest [LO1] First City Bank pays 8 percent simple interest on its savings account

balances, whereas Second City Bank pays 8 percent interest compounded annually. If you made a $68,000 deposit in each bank, how much more money would you earn from your Second City Bank account at the end of 8 years
Business
1 answer:
Anna71 [15]2 years ago
7 0

Answer:

$14,343.25

Explanation:

The computation is shown below;

For the first bank

The value of investment is

= $68,000 × 8% × 8 + $68,000

= $111,520

For the second bank

= $68,000 × (1 + 0.08)^8

= $125,863.25

So, the difference in these both amount should be

= $125,863.25 - $111,520

= $14,343.25

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When bonds are issued at their face amount, the journal entry will include a __________ to __________.
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D. Credit; bonds payable
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Jerrod is relatively new to Xenon Corporation and wants to make sure that he makes a good impression on his coworkers and superv
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On May 23, Stoltz Realty Inc. issued for cash 80,000 shares of no-par common stock (with a stated value of $3) at $12. On July 6
Slav-nsk [51]

Answer:

23rd May

Dr Cash                                                          960,000

Cr Common stock                                        240,000

Cr Paid-in Capital - Common Stock            720,000

( to record the issuance of 80,000 common shares for cash)

6th July

Dr Cash                           900,000

Cr Preferred stock          900,000

( to record the issuance of 18,000 preferred shares for cash)

15th September

Dr Cash                                                   750,000

Cr Common stock                                  150,000

Cr Paid-in capital - Common Stock       600,000

( to record the issuance of 50,000 common shares for cash)

Explanation:

Working notes for each transactions:

* 23rd May:

Cash increases by: Amount of stocks issued * Price at issuance = 80,000 * 12 = $960,000

Common stock account increases by: Amount of stock issued * Stated value = 80,000 * 3 = 240,000

Paid-in capital account increased by: Amount of stock issued * ( Price at issuance - Stated value) = 80,000 * 9 = $720,000

* 6th July:

Cash increases by: Amount of stocks issued * Price at issuance = 18,000 * 50 = $900,000

Preferred stock account increases by: Amount of stock issued * Par value = 18,000 * 50 = $900,000;

As shares are issued at par; no paid-in capital amount recorded.

* 15th September:

Cash increases by: Amount of stocks issued * Price at issuance = 50,000 * 15 = $750,000

Common stock account increases by: Amount of stock issued * Stated value = 50,000 * 3 = 150,000

Paid-in capital account increased by: Amount of stock issued * ( Price at issuance - Stated value) = 50,000 * 12 = $600,000.

3 0
3 years ago
The group responsible for establishing standards that identify material specifications and testing procedures is ____.
Mrrafil [7]

Answer: ASTM International

Explanation:

The group that is responsible for the establishment of standards which identify material specifications and testing procedures is referred to as ASTM International.

ASTM International was formerly called the American Society for Testing and Materials. This is an international standards organization which technical standards for different products, materials, systems, and services.

It was put in place in order to improve safety and to also improve the quality o both the consumer and the industrial products.

6 0
3 years ago
A body in the solar system has a period of 10,759.22 days and a perihelion speed of 10.18 km/s. a. Calculate the aphelion radius
anastassius [24]

Answer:

Explanation:

From the information given, by applying Kepler's 3rd law,

T^2 \alpha  a^3

where;

T = period

a = semi major axis

T = 356 days (for earth)

a = 1 AU = 1.496 \times 10^8 \ km

Therefore, T^2 = ca^3

c= \dfrac{365^2}{(1.496 \times 10^8)^3}

c = 3.9791 \times 10^{20} \ day^2/km^3

However, if the body in the solar system has a period of 10.759.22 days, then, a =?

∴

T^2 = ca^3

a3 = \dfrac{10759.22^2}{3.9791 \times 10^{-20}}

a^3 = 2.9092 \times 10^{27}

a= \sqrt[3]{2.9092 \times 10^{27}}

a = 1.4275 \times 10^9 \ km

However, the velocity for a perihelion = 10.18 km/s

Using the formula

v = \sqrt{GM ( \dfrac{2}{r}-\dfrac{1}{a})} to calculate the radius, we have:

G = 6.674 \times 10^{-11}

M = 1.989\times 10^{30} \ kg

r = perihelion

v ^2= GM ( \dfrac{2}{r}-\dfrac{1}{a})

(10.18 \times 10^3) ^2= 6.674 \times 10^{-11} \times 1.989 \times 10^{30}  ( \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}})

7.8068 \times 10^{-13}= \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}}

\dfrac{2}{r} = 1.4824 \times 10^{-12}

r = \dfrac{2}{1.4824 \times 10^{-12}}

r = 1.349 \times 10^{12}

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity

∴

1.349 \times 10^{12} = 1.425 \times 10^{12} ( 1 - e)

1.349 \times 10^{12}  -  1.425 \times 10^{12}= -  1.425 \times 10^{12} (e)

-7.6\times 10^{10}= -  1.425 \times 10^{12} (e)

\dfrac{-7.6\times 10^{10}}{-  1.425 \times 10^{12}}=  (e)

e ( eccentricity) = 0.0533

Aphelion radius in natural miles, r = a( 1+ e)

r = 1.425 \times 10^{12} ( 1 + 0.0533)

r = 1.50 \times 10^{12} \ m

to nautical miles, we have:

r = 1.50 \times 10^{12} \times 0.00054  \ nautical \ mile

radius of aphelion \mathbf{r = 8.10 \times 10^8} nautical miles

In respect to the value of a( i.e 1.4275 \times  10^9 \ km)

the body of the solar system is Saturn

5 0
3 years ago
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