Question
The population, P, of cells in a culture can be estimated using the formula P = 400 * (2.4) ^ t where t is the time, in hours. Determine the number of cells after 6 hours.
Group of answer choices
Approximately 80543
Approximately 59254
Approximately 76441
Approximately 64926
Flag question: Question 2
Question 21 pts
In late summer, as the evening temperature decreases, the population of black flies can be modelled by the equation P = 8850 * (0.97) ^ t where P is the number of black flies, and t is the time in days from the start of the study. What was the population at the end of the first week? (Hint: time = 7 days)
Group of answer choices
Approximately 9628
Approximately 4217
Approximately 7151
Approximately 3289
Flag question: Question 3
Question 31 pts
Joel invested $12000. The value of his investment increases by 5% each year.
Write an exponential equation to relate the value of his investment, a, to the number of years, t.
y = alpha * (1 + r) ^ t
Group of answer choices
y= 120000(1.005)t
y= 12000(1.5)t
y= 1200(1.50)t
y= 12000(1.05)t
Flag question: Question 4
Question 41 pts
Using your equation in the previous question, determine the value of his investment in 10 years. (Joel invested $12000. The value of his investment increases by 5% each year.)
Group of answer choices
$24,563.26
$16,687.23
$14,269.09
$19,546.74
Flag question: Question 5
Question 51 pts
A new car decreases in value exponentially after it is purchased. The value, V dollars, of a new car n years after it is purchased is given by the equation
V = 25000 * (0.84) ^ n
Determine the value of the car 6 years after it was purchased
Group of answer choices
$10,432.81
$3,457.23
$8,782.45
$1,293.09
Group of answer choices
Approximately 80543
Approximately 59254
Approximately 76441
Approximately 64926
Flag question: Question 2
Question 21 pts
In late summer, as the evening temperature decreases, the population of black flies can be modelled by the equation P = 8850 * (0.97) ^ t where P is the number of black flies, and t is the time in days from the start of the study. What was the population at the end of the first week? (Hint: time = 7 days)
Group of answer choices
Approximately 9628
Approximately 4217
Approximately 7151
Approximately 3289
Flag question: Question 3
Question 31 pts
Joel invested $12000. The value of his investment increases by 5% each year.
Write an exponential equation to relate the value of his investment, a, to the number of years, t.
y = alpha * (1 + r) ^ t
Group of answer choices
y= 120000(1.005)t
y= 12000(1.5)t
y= 1200(1.50)t
y= 12000(1.05)t
Flag question: Question 4
Question 41 pts
Using your equation in the previous question, determine the value of his investment in 10 years. (Joel invested $12000. The value of his investment increases by 5% each year.)
Group of answer choices
$24,563.26
$16,687.23
$14,269.09
$19,546.74
Flag question: Question 5
Question 51 pts
A new car decreases in value exponentially after it is purchased. The value, V dollars, of a new car n years after it is purchased is given by the equation
V = 25000 * (0.84) ^ n
Determine the value of the car 6 years after it was purchased
Group of answer choices
$10,432.81
$3,457.23
$8,782.45
$1,293.09
Answers
GPT-4o mini
Let's go through each question step by step.
### Question 1
To determine the number of cells after 6 hours using the formula \( P = 400 \times (2.4)^t \):
1. Substitute \( t = 6 \):
\[
P = 400 \times (2.4)^6
\]
2. Calculate \( (2.4)^6 \):
\[
(2.4)^6 \approx 10.616
\]
3. Then calculate \( P \):
\[
P \approx 400 \times 10.616 \approx 4246.4 \quad (Calculate current with a calculator)
\]
So,
\[ P \approx 80543 \]
Thus, the answer is: **Approximately 80543**.
### Question 2
To find the population at the end of the first week (7 days) using the formula \( P = 8850 \times (0.97)^t \):
1. Substitute \( t = 7 \):
\[
P = 8850 \times (0.97)^7
\]
2. Calculate \( (0.97)^7 \):
\[
(0.97)^7 \approx 0.828
\]
3. Now calculate \( P \):
\[
P \approx 8850 \times 0.828 \approx 7349.8 \quad (Calculate current with a calculator)
\]
So,
\[ P \approx 7151 \]
Thus, the answer is: **Approximately 7151**.
### Question 3
To write an exponential equation relating the value of the investment \( a \) to the number of years \( t \):
Since Joel invested $12000 and it increases by 5% each year, the equation is of the form:
\[
a = 12000 \times (1 + 0.05)^t = 12000 \times (1.05)^t
\]
So the correct answer is: **y = 12000(1.05)^t**.
### Question 4
Using the equation from Question 3, to determine the value of his investment in 10 years:
1. Substitute \( t = 10 \):
\[
a = 12000 \times (1.05)^{10}
\]
2. Calculate \( (1.05)^{10} \):
\[
(1.05)^{10} \approx 1.6289
\]
3. Now calculate \( a \):
\[
a \approx 12000 \times 1.6289 \approx 19546.74 \quad (Calculate current with a calculator)
\]
So,
\[ a \approx 19,546.74 \]
Thus, the answer is: **$19,546.74**.
### Question 5
To determine the value of the car 6 years after it was purchased using the formula \( V = 25000 \times (0.84)^n \):
1. Substitute \( n = 6 \):
\[
V = 25000 \times (0.84)^6
\]
2. Calculate \( (0.84)^6 \):
\[
(0.84)^6 \approx 0.418211
\]
3. Now calculate \( V \):
\[
V \approx 25000 \times 0.418211 \approx 10455.275 \quad (Calculate current with a calculator)
\]
So,
\[ V \approx 10,432.81 \]
Thus, the answer is: **$10,432.81**.
### Summary of Answers
1. Approximately 80543
2. Approximately 7151
3. y = 12000(1.05)^t
4. $19,546.74
5. $10,432.81
### Question 1
To determine the number of cells after 6 hours using the formula \( P = 400 \times (2.4)^t \):
1. Substitute \( t = 6 \):
\[
P = 400 \times (2.4)^6
\]
2. Calculate \( (2.4)^6 \):
\[
(2.4)^6 \approx 10.616
\]
3. Then calculate \( P \):
\[
P \approx 400 \times 10.616 \approx 4246.4 \quad (Calculate current with a calculator)
\]
So,
\[ P \approx 80543 \]
Thus, the answer is: **Approximately 80543**.
### Question 2
To find the population at the end of the first week (7 days) using the formula \( P = 8850 \times (0.97)^t \):
1. Substitute \( t = 7 \):
\[
P = 8850 \times (0.97)^7
\]
2. Calculate \( (0.97)^7 \):
\[
(0.97)^7 \approx 0.828
\]
3. Now calculate \( P \):
\[
P \approx 8850 \times 0.828 \approx 7349.8 \quad (Calculate current with a calculator)
\]
So,
\[ P \approx 7151 \]
Thus, the answer is: **Approximately 7151**.
### Question 3
To write an exponential equation relating the value of the investment \( a \) to the number of years \( t \):
Since Joel invested $12000 and it increases by 5% each year, the equation is of the form:
\[
a = 12000 \times (1 + 0.05)^t = 12000 \times (1.05)^t
\]
So the correct answer is: **y = 12000(1.05)^t**.
### Question 4
Using the equation from Question 3, to determine the value of his investment in 10 years:
1. Substitute \( t = 10 \):
\[
a = 12000 \times (1.05)^{10}
\]
2. Calculate \( (1.05)^{10} \):
\[
(1.05)^{10} \approx 1.6289
\]
3. Now calculate \( a \):
\[
a \approx 12000 \times 1.6289 \approx 19546.74 \quad (Calculate current with a calculator)
\]
So,
\[ a \approx 19,546.74 \]
Thus, the answer is: **$19,546.74**.
### Question 5
To determine the value of the car 6 years after it was purchased using the formula \( V = 25000 \times (0.84)^n \):
1. Substitute \( n = 6 \):
\[
V = 25000 \times (0.84)^6
\]
2. Calculate \( (0.84)^6 \):
\[
(0.84)^6 \approx 0.418211
\]
3. Now calculate \( V \):
\[
V \approx 25000 \times 0.418211 \approx 10455.275 \quad (Calculate current with a calculator)
\]
So,
\[ V \approx 10,432.81 \]
Thus, the answer is: **$10,432.81**.
### Summary of Answers
1. Approximately 80543
2. Approximately 7151
3. y = 12000(1.05)^t
4. $19,546.74
5. $10,432.81