Which on these time measurements is the smallest?

Usimov [2.4K]

Answer:

Part a. t = 7.29 years.

Part b. t = 27.73 years.

Part c. p = $3894.00

Step-by-step explanation:

The formula for continuous compounding is: A = p*e^(rt); where A is the amount after compounding, p is the principle, e is the mathematical constant (2.718281), r is the rate of interest, and t is the time in years.

Part a. It is given that p = $2000, r = 2.5%, and A = $2400. In this part, t is unknown. Therefore: 2400 = 2000*e^(2.5t). This implies 1.2 = e^(0.025t). Taking natural logarithm on both sides yields ln(1.2) = ln(e^(0.025t)). A logarithmic property is that the power of the logarithmic expression can be shifted on the left side of the whole expression, thus multiplying it with the expression. Therefore, ln(1.2) = 0.025t*ln(e). Since ln(e) = 1, and making t the subject gives t = ln(1.2)/0.025. This means that t = 7.29 years (rounded to the nearest 2 decimal places)!!!

Part b. It is given that p = $2000, r = 2.5%, and A = $4000. In this part, t is unknown. Therefore: 4000 = 2000*e^(2.5t). This implies 2 = e^(0.025t). Taking natural logarithm on both sides yields ln(2) = ln(e^(0.025t)). A logarithmic property is that the power of the logarithmic expression can be shifted on the left side of the whole expression, thus multiplying it with the expression. Therefore, ln(2) = 0.025t*ln(e). Since ln(e) = 1, and making t the subject gives t = ln(2)/0.025. This means that t = 27.73 years (rounded to the nearest 2 decimal places)!!!

Part c. It is given that A = $5000, r = 2.5%, and t = 10 years. In this part, p is unknown. Therefore 5000 = p*e^(0.025*10). This implies 5000 = p*e^(0.25). Making p the subject gives p = 5000/e^0.25. This means that p = $3894.00(rounded to the nearest 2 decimal places)!!!

4 0

3 years ago

Write the linear equation that gives the rule for this table.

Maksim231197 [3]

to get the equation of any straight line all we need is two points, well, let's just grab them from the table.

hmmm let's use (-49 , -39) and hmmm say (67 , 77)

$(\stackrel{x_1}{-49}~,~\stackrel{y_1}{-39})\qquad (\stackrel{x_2}{67}~,~\stackrel{y_2}{77}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{77}-\stackrel{y1}{(-39)}}}{\underset{run} {\underset{x_2}{67}-\underset{x_1}{(-49)}}}\implies \cfrac{77+39}{67+49}\implies \cfrac{116}{116}\implies 1$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-39)}=\stackrel{m}{1}(x-\stackrel{x_1}{(-49)}) \\\\\\ y+39=1(x+49)\implies y+39=x+49\implies y=x+10$