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)!!!
Answer:
25 one-dollar coins, 16 half-dollar coins, and 164 quarters
Step-by-step explanation:
First, set up equations based on the information given:
Then, substitute <em>q</em> in the first equation with the expression from the third equation:
![0.25[4(d+h)]+0.50h+1.00d=74\\1d+1h+0.50h+1.00d=74\\2d+1.5h=74](https://tex.z-dn.net/?f=0.25%5B4%28d%2Bh%29%5D%2B0.50h%2B1.00d%3D74%5C%5C1d%2B1h%2B0.50h%2B1.00d%3D74%5C%5C2d%2B1.5h%3D74)
Next, substitute <em>h</em> in that equation with the expression from the second equation:
Solve for <em>d</em>, the number of one-dollar coins:
Substitute 25 for <em>d</em> in the second equation to find <em>h</em>, the number of half-dollar coins:
Substitute 25 for <em>d</em> and 16 for <em>h</em> in the third equation to find <em>q</em>, the number of quarters:
Then, verify that the coins total $74:
Next, verify that the number of half-dollar coins is one more than three-fifths of the number of one-dollar coins:
Finally, verify that the number of quarters is four times the number one-dollar and half-dollar coins together:
The answer is false, it is only true for postulates, not conjectures.
-9y-42x
Because they are like terms that is the lowest they can go
Sample space for all possible outcomes:
HH, HT, TH, TT
Sample space for event where heads is the first toss:
HT, HH
