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3 years ago

5

Bawl purchased ABC bonds on 1/1/21. Data regarding these available-for-sale securities follow: Cost MV December 31, 2021 $100,00

0 $ 91,000 December 31, 2022 100,000 111,000 December 31, 2023 100,000 106,000 The unrealized Gain/Loss reported in OCI of the 2023 Comprehensive Income statement is:

1 answer:

GrogVix [38]3 years ago

6 0

Answer:

Bawl with ABC bonds

The unrealized Gain/Loss reported in OCI of the 2023 Comprehensive Income statement is:

A Loss of $5,000

Explanation:

a) Data and Calculations:

Cost MV Unrealized Profit or (Loss)

December 31, 2021 $100,000 $ 91,000 $9,000 (Loss)

December 31, 2022 100,000 111,000 20,000

December 31, 2023 100,000 106,000 5,000 (Loss)

Available-for-sale Investment

Debit Credit

Dec 31 100,000

Loss 9,000

Dec 31 91,000

Profit 20,000

Dec 31 111,000

Loss 5,000

Dec 31 106,000

The Available-for-sale Investment will show a loss of $5,000 in the Other Comprehensive Income of the 2023 Comprehensive Income Statement based on the yearly adjustments to the account with losses and profits.

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Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg, and 5.6 cm of wa

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Answer:

$1170839.28 dyn/cm^2$

16.9816 psia

$117.083928 kN/m^2$

Explanation:

To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.

$P_{abs}=P_{atm}+P_G$

And the gauge pressure is given by the contributions of columns of water ( $P_{w}$ ) and mercury( $P_{Hg}$ ), we can calculate the contribution of each column as:

$P= \rho g h$ (*)

where $\rho$ is the respective density, g gravity and h is height.

So we have all the data required to use the above equations ( $P_{atm}$ , height and density of each column) we only need to be carefully with the units.

For simplicity we can to express all pressure contributions in mmHg ( $P_{atm}$ , $P_{w}$ and $P_{Hg}$ ). Note that the units "x" mmHg means the pressure at the bottom of a column of mercury of "x" mm high. For example, in this case we have a column 12.1 cm of Hg, that is a column of 121 mmHg (passing from cm to mm only requires multiply by 10) pressure exerted by that column is 121 mmHg.

Now pressure of 5.6 cm (56 mm) of water would be 56 mm of water, but it is not the same that mmHg, since the density of water is lower, the pressure exerted by 1 mm of water is lower than the exerted by 1 mm of Hg. The conversion between mmHg and mm of water is given by the relation between the densities.

$mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}$

$mmHg=\frac{0.998*mmH_2O}{13.55}=0.0737 mmH_2O$

And pressure of water in mmHg is

$0.0737*56=4.1246 mmHg$

The absolute pressure is:

$P_{abs}=P_{atm}+P_G= 756 + 121 + 4.1246 = 881.1246 mmHg = 88.11246cmHg$

To pass to dyn/cm^2 units we need to use the equation (*)

$P= \rho g h = 13.55 \frac{g}{cm^3} * 980.665 \frac{cm}{s^2} * 88.11246 cmHg = 1170839.28 \frac{g}{cm s^2} = 1170839.28 \frac{dyn}{cm^2}$

Note: We need to use cm Hg for units coherence

Now passing from dyn/ $cm^2$ to kN/ $m^2$ (or kPa) we need to consider that 1 dyn is $10^{-8}$ kN and 1 $cm^2$ is $10^{-4} m^2$ .

$1170839.28 \frac{dyn}{cm^2} * \frac{10^{-8}kN}{1 dyn}*\frac{cm^2}{10^{-4}m^2}=117.083928kN/m^2$

Now passing kN/ $m^2$ to psia. We need to consider that 1 psia is 6.89476.

$117.083928kN/m^2*\frac{1psia}{6.89476kN/m^2}=16.9816 psia$

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Answer:

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frozen [14]

Answer:

Requirement 1

<em>High Point = March</em>

<em>Low Point = January</em>

Requirement 2

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Requirement 3

<em>Total Cost = $1,457 + $0.41 × Activity Level</em>

Requirement 4

<em>Total predicted cost for September $ 2,482</em>

<em>Total fixed cost for September $1,457</em>

<em>Total predicted variable cost for September $1,025</em>

Explanation:

Requirement 1

High Point = March

Low Point = January

Requirement 2

<em>Variable Cost = Difference in Overhead Cost (High - Low)/ Difference in Activity (High - Low)</em>

= $2,810 - $1,744 / 3,300 - 700

= $1,066/ 2,600

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Lets Select, January 700 $1,744

Fixed Cost Component = Total Overhead Cost - Variable Cost × Selected Activity.

= $1,744 - $0.41 × 700

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Cost formula for tanning services

<em>Total Cost = Fixed Cost + Variable Cost</em>

thus,

Total Cost = $1,457 + $0.41 × Activity Level

Requirement 4

Total Cost = $1,457 + $0.41 × Activity Level

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