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1 year ago

14

Raul deposited $3000 into a bank account that earned simple interest each year. After 3. 5 years, he had earned $262. 50 in inte

rest. No money was deposited into or withdrawn from the account. What was the annual interest rate?

1 answer:

lions [1.4K]1 year ago

8 0

Answer:

262.50-3000=222.50

therefore 3000×100/222.50×3.5=$4,719

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A private pilot wishes to insure his airplane for $. The insurance company estimates that a total loss will occur with probabil

stealth61 [152]

The premium that the insurance company should charge each year to realize an average profit of $500 is $6,900.

First step is to calculated the expected amount to pay

Expected amount=Total loss +50% loss+25% loss

Expected amount=$200,000(0.002)(1)+$200,000(0.01)(0.5)+$200,000(0.1)(0.25)

Expected amount=$400+$1,000+$5,000

Expected amount=$6,400

Second step is to calculate the premium

Premium=Expected amount+ Average profit

Premium=$6,400+$500

Premium=$6,900

Inconclusion the premium that the insurance company should charge each year to realize an average profit of $500 is $6,900.

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2 years ago

For the galvanic (voltaic) cell fe(s) + mn²⁺(aq) ⟶ fe²⁺(aq) + mn(s) (e° = 0.77 v at 25°c), what is [fe²⁺] if [mn²⁺] = 0.050 m an

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The concentration of Iron in the galvanic (voltaic) cell Fe(s) + Mn²⁺(aq) ⟶ Fe²⁺(aq) + Mn(s) is 0.02297 M.

<h3>What is the Nernst Equation?</h3>

The Nernst equation enables us to identify the cell potential(voltage) in presence of non-standard conditions in a galvanic cell. It can be expressed by using the formula:

$\mathbf{E_{cell} = E_o - \dfrac{0.059}{n} \times log \dfrac{[Fe^+]}{[Mn^{2+}]}}$

where;

n = Number of electrons = 2
$\mathbf{E_o}$ = Initial voltage = 0.77 V
$\mathbf{E_{cell}}$ = Cell voltage = 0.78 V
$\mathbf{[Mn^{2+}]}$ = Manganese concentration = 0.050 M

Replacing the values into the above equation, we have:

$\mathbf{0.78 = 0.77 - \dfrac{0.059}{2} \times log \dfrac{[Fe^{2+}]}{[0.050]}}$

$\mathbf{0.78 -0.77= -0.0296\times log \dfrac{[Fe^{2+}]}{[0.050]}}$

$\mathbf{log^{-1} (-0.3378) = \dfrac{[Fe^{2+}]}{[0.050]}}$

$\mathbf{Fe^{2+} = (0.4594\times 0.050 )\ M}$

$\mathbf{Fe^{2+} =0.02297 \ M }$

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