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Taya2010 [7]
2 years ago
8

A stock sells for $6.99 on December 31, providing the seller with a 6% annual return. What was the price of the stock at the beg

inning of the year? $6.59 $1.16 $7.42 $5.84
Business
1 answer:
Dimas [21]2 years ago
6 0

Answer:

Correct option is 6.59

Explanation:

Selling price of stock at the end of the year is $6.99. Annual return rate is 6%. Price of stock at the beginning will be present value of stock valued at the end discounted at 6%. Computation is as shown below:

Present\ value\ or\ price\ of\ stock = Selling\ price\left ( \frac{1}{1+i} \right )^{n}

= 6.99\left ( \frac{1}{1+0.06} \right )^{1}

= \frac{6.99}{1.06}

= $6.59

Therefore, Stock's price in the beginning of the year is $6.59.

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Pippin Inc. has provided the following data concerning one of the products in its standard cost system. Variable manufacturing o
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Answer:

920 (Unfavorable)

Explanation:

Labor rate variance = Actual direct labor hours (Actual direct labor rate - Standard direct labor rate)

Labor rate variance = 2,300 * ($21.7 - $21.3)

Labor rate variance = 2,300 * 0.4

Labor rate variance = 920 (Unfavorable)

6 0
3 years ago
<11{6[1(07)71]}> The higher the price of an antique, the greater people’s expectation that the object is rare. That, in tu
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a. should be discouraged because it lessens a quality that makes that antique desirable

Explanation:

In pricing theory, the price for a good or service should increase as its scarcity increases. Now selling the antique at a bargain price will reduce the price of it and thereby making it less scarce and rare.

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Read 2 more answers
Let $S$ be the set of complex numbers of the form $a + bi,$ where $a$ and $b$ are integers. We say that $z \in S$ is a unit if t
lakkis [162]

Answer:

Number of units possible in S are 4.

Explanation:

Given <em>S</em> is a set of complex number of the form a+bi where <em>a</em> and <em>b</em> are integers.

z\in S is a unit if w\in z exists such that zw=1.

To find:

Number of units possible = ?

Solution:

Given that:

zw = 1

Taking modulus both sides:

|zw| = |1|

Using the property that modulus of product of two complex numbers is equal to their individual modulus multiplied.

i.e.

|z_1z_2|=|z_1|.|z_2|

So,

|zw| = |1|\\\Rightarrow |zw| =|z|.|w| =1\\\Rightarrow |z|=\dfrac{1}{|w|}......... (1)

Let z=a+bi

Then modulus of z is   |z| = \sqrt{a^2+b^2}

Given that a and b are <em>integers</em>, so the equation (1) can be true only when |z| = |w| =1 (Reciprocal of 1 is 1). Modulus can be equal only when one of the following is satisfied:

(a = 1, b = 0) ,  (a = -1, b = 0), (a = 0, b = 1) OR (a = 0, b = -1)

So, the possible complex numbers can be:

1.\ 1 + 0i = 1\\2.\ -1 + 0i = -1\\3.\ 0+ 1i = i\\4.\ 0 -1i = -i

Hence, number of units possible in S are 4.

6 0
3 years ago
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His side gig was the equivalent of self contracting. He would definitely have to report of he made over $400.
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