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erastova
2 months ago
7

If the FDIC has an insurance fund of $67.8 billion and must use 7.6% of it to cover several failed banks, approximately how much

money is left in the fund?
a.
$5,153 million
b.
$57.49 billion
c.
$72.95 billion
d.
$62.65 billion
Business
2 answers:
Free_Kalibri [3.7K]2 months ago
7 0

<span>If the FDIC possesses an insurance fund of $67.8 billion and needs to allocate 7.6% towards covering several failed banks, how much funds remain in the reserve? <span>Approximately $62.65 billion remains in the fund. </span></span>

To calculate:

Approximate remaining funds = ($67.8 billion)(0.076) = $5.15 billion

<span>Approximate remaining funds = $67.8 billion - $5.15 billion </span>

<span>Approximate remaining funds = $62.65 billion</span>

harina [3.8K]2 months ago
6 0
The accurate answer among the options is d. $62.65 billion. This indicates the approximate funds left in the fund after the 7.6% deduction for covering several banks that failed. Thank you for your inquiry. I hope this information was useful. Please reach out if you need further assistance. 
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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:

16.9816 psia

117.083928 kN/m^2

Explanation:

To determine the absolute pressure at the tube's bottom, we must add atmospheric pressure to gauge pressure.

P_{abs}=P_{atm}+P_G

Gauge pressure accounts for contributions from water columns (P_{w}) and mercury (P_{Hg}), allowing us to derive the contribution of each as:

P= \rho g h (*)

where \rho represents density, g is gravity, and h is height.

With all necessary data available to apply the above equations (P_{atm}, height, and density of each fluid), we must be diligent about unit consistency.

For clarity, we can express all pressure contributions in mmHg ( P_{atm}, P_{w}, and P_{Hg}). The units "x" mmHg indicate the pressure at the bottom of a mercury column that is "x" mm tall. In this case, a 12.1 cm Hg column equals 121 mmHg (conversion from cm to mm requires multiplying by 10) showing the pressure exerted is 121 mmHg.

For a water pressure at 5.6 cm (56 mm), it equals 56 mm of water. However, this differs from mmHg because water's density is less than mercury's, resulting in 1 mm of water exerting less pressure than 1 mm of Hg. The conversion between mmHg and mm of water relies on their densities.

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

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

Thus, water pressure in mmHg is calculated as

0.0737*56=4.1246 mmHg

The absolute pressure is computed as:

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

To convert to dyn/cm^2 units, we will utilize 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: It is essential to maintain cm Hg for uniformity.

Next, to convert from dyn/cm^2 to kN/m^2 (or kPa), we must remember that 1 dyn equals 10^{-8} kN and that 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

Finally, transitioning from kN/m^2 to psia, we must take into account that 1 psia equals 6.89476.

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

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