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I am Lyosha [343]

3 years ago

8

Suppose an annuity pays 6% annual interest, compounded semi-annually. You invest in this annuity by contributing $4,500 semiannu

ally for 6 years. What will the annuity be worth after 6 years?

1 answer:

yaroslaw [1]3 years ago

5 0

Answer:

$3240

Step-by-step explanation:

hope it is well understood

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Answer:

4u+4

Step-by-step explanation:

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3 years ago

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Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)

Crazy boy [7]

Answer:

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Step-by-step explanation:

To find : Convert $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}$ into $\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Solution :

We convert units one by one,

$1\text{ m}^2=10.7639\text{ ft}^2$

$1\text{ sec}=\frac{1}{3600}\text{ hour}$

$1\text{ cal}=0.003968\text{ BTU}$

Converting temperature unit,

$^\circ C\times \frac{9}{5}+32=^\circ F$

$1^\circ C\times \frac{9}{5}+32=33.8^\circ F$

So, $1^\circ C=33.8^\circ F$

Substitute all the values in the unit conversion,

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Therefore, The conversion of unit is $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

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3 years ago

Modeling Compound Interest In Exercise, complete the table for an account in which interest is compounded continuously.

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Answer:

This problem requires us to calculate, the value of investment after 10 and 25 years, and also tell the time after which intial investment amount will double. Investment rate and initial investment amount is given in the question.

Value of investment after 10 year = 600(1+8%)^10 = $ 1,295

Value of investment after 25 year = 600(1+8%)^25 = $ 4,109

Time after which investment amount double (n)

1200 = 600 (1.08)^n

Log 2 = n log 1.08

n = 9 years

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