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3 years ago

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Chris bought Timberland boots because he felt they were perfect for his outdoor activities. Pat bought the same kind of boots be

cause he felt they were stylish, especially with the logo clearly visible. The psychological factor driving Pat's behavior is ___________.b. his evoked set.
c. lifestyle. d. learned behavior.
e. brand association.

1 answer:

frutty [35]3 years ago

5 0

Answer:

c. lifestyle

Explanation:

The reason why Patrick made his choice was because of his style of living, the way he spends his time and money. He is someone that love to live a stylish life, thus the reason behind picking the boot. This better reveals psychology and not functional needs.

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In mary poppins, what animal was on the end of mary poppins' umbrella that spoke?

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Pretty sure it was a parrot.

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In the United States, the money supply is determined: A) only by the Fed. B) only by the behavior of individuals who hold money

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Answer: Joint by the FED and by the behavior of individuals who hold money and of banks which money is held.

Explanation: The Federal Reserve System, often referred as the Federal reserve or simply "the fed", is the central bank of the united states. It was created by the congress to provide the nation with a safer, more flexible, and more stable monetary and financial system. The FED was created on December 23, 1913, when president Woodrow Wilson signed the FEDERAL RESERVE ACT into law. The Fed and the behavior of individuals not only define how much money are available, they can also define macroeconomic indicators like inflation.

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If the fed wants to _______ the money supply, it discourages bank loans.

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3 years ago

Meng Co. maintains a $300 petty cash fund. On January 31, the fund is replenished. The accumulated receipts on that date represe

Alona [7]

Answer: The correct answer is e) $32.

Explanation:

Petty cash fund. $300

Office supplies. (80)

Merchandise inventory. (160)

Miscellaneous expenses. (20)

Cash shortage. (8)

Balance in petty cash. $32

In terms of accounting entries,

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Debit Merchandise inventory. $160

Debit Miscellaneous expenses. $20

Debit Cash shortage. $8

Credit Petty cash refund. $268

In the above entries, $268 would be refunded to petty cash fund to reinstate it to $300.

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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

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3 years ago