3)the new york mets sign a new player for $8,000,000 and his salary goes up by 3% every year. (a)initial value: (b) growth factor: (c) equation: 4) a certain stock was worth $42 at the beginning of the day. every hour the stock goes down by 15%. (a)initial value: (b) growth factor: (c) equation: word problem 1: a flu outbreak hits your school on monday, with an initial number of 20 ill students coming to school. the number of ill students increases by 25% per hour. (a) this situation is an example of exponential. (b) create a function to models this monday flu outbreak. (c) how many students will be ill after 6 hours? word problem 2: in 2000, the world population was about 6.09 billion. during the next 13 years, the world population increased by about 18% each year. (a) write an exponential growth model giving the population y (in billions) t years after 2000. estimate the world population in 2005. (b) estimate the year when the world population was 7 billion. error analysis you invest $500 in the stock of a company. the value of the stock decreases 2% each year. describe and correct the error in writing a model for the value of the stock after t years. y=(initial amount)(decay factor)^t y=500(0.02)^t closing: write a summary of todays lesson in your own words.

Question

Accepted Answer

### Problem 3:
# Explanation:
## Step1: Identify Initial Value
The initial salary is $8,000,000.
## Step2: Determine Growth Factor
A 3% increase means the growth factor is $1 + 0.03 = 1.03$.
## Step3: Formulate the Equation
Using the exponential growth formula $y = a(1 + r)^t$, where $a = 8000000$ and $r = 0.03$, the equation is $y = 8000000(1.03)^t$.
# Answer:
(A) Initial value: $\boldsymbol{8000000}$
(B) Growth factor: $\boldsymbol{1.03}$
(C) Equation: $\boldsymbol{y = 8000000(1.03)^t}$

### Problem 4:
# Explanation:
## Step1: Identify Initial Value
The initial stock value is $42.
## Step2: Determine Growth Factor (Decay Factor)
A 15% decrease means the decay factor is $1 - 0.15 = 0.85$.
## Step3: Formulate the Equation
Using the exponential decay formula $y = a(1 - r)^t$, where $a = 42$ and $r = 0.15$, the equation is $y = 42(0.85)^t$ (where $t$ is in hours).
# Answer:
(A) Initial value: $\boldsymbol{42}$
(B) Growth factor (Decay factor): $\boldsymbol{0.85}$
(C) Equation: $\boldsymbol{y = 42(0.85)^t}$

### Word Problem 1:
#### (A)
# Brief Explanations:
The number of ill students is increasing by a percentage each hour, so it's exponential growth.
# Answer:
(A) This situation is an example of exponential $\boldsymbol{growth}$.

#### (B)
# Explanation:
## Step1: Identify Initial Value and Growth Factor
Initial number of ill students $a = 20$, growth rate $r = 0.25$, so growth factor is $1 + 0.25 = 1.25$.
## Step2: Formulate the Function
Using $y = a(1 + r)^t$, the function is $y = 20(1.25)^t$ (where $t$ is in hours).
# Answer:
(B) Function: $\boldsymbol{y = 20(1.25)^t}$

#### (C)
# Explanation:
## Step1: Substitute $t = 6$ into the Function
$y = 20(1.25)^6$
## Step2: Calculate $(1.25)^6$
$1.25^6 \approx 2.44140625$
## Step3: Multiply by Initial Value
$y = 20 \times 2.44140625 \approx 48.828125$, so approximately 49 students (or keep as a decimal).
# Answer:
(C) After 6 hours, approximately $\boldsymbol{49}$ students (or $\boldsymbol{48.83}$ if decimal is preferred).

### Word Problem 2:
#### (A)
# Explanation:
## Step1: Identify Initial Value and Growth Factor
Initial population $a = 6.09$ billion, growth rate $r = 0.18$, so growth factor is $1 + 0.18 = 1.18$.
## Step2: Formulate the Model
The exponential growth model is $y = 6.09(1.18)^t$. For 2005, $t = 2005 - 2000 = 5$.
## Step3: Calculate Population in 2005
$y = 6.09(1.18)^5$. Calculate $1.18^5 \approx 2.287757$, then $y \approx 6.09 \times 2.287757 \approx 13.93$ billion.
# Answer:
(A) Model: $\boldsymbol{y = 6.09(1.18)^t}$; Population in 2005: ~$\boldsymbol{13.93}$ billion.

#### (B)
# Explanation:
## Step1: Set Up the Equation
$7 = 6.09(1.18)^t$
## Step2: Solve for $t$
Divide both sides by 6.09: $\frac{7}{6.09} = (1.18)^t$
Take natural log: $\ln(\frac{7}{6.09}) = t\ln(1.18)$
Calculate $\ln(\frac{7}{6.09}) \approx \ln(1.1494) \approx 0.139$, $\ln(1.18) \approx 0.1655$
$t \approx \frac{0.139}{0.1655} \approx 0.84$ years. So around 2000 + 1 = 2001 (since $t \approx 0.84$ is close to 1).
# Answer:
(B) Estimated year: ~$\boldsymbol{2001}$

### ERROR ANALYSIS:
# Brief Explanations:
The decay factor was incorrect. A 2% decrease means the remaining value is $1 - 0.02 = 0.98$, not 0.02. The correct model is $y = 500(0.98)^t$.
# Answer:
Error: Used decay factor 0.02 (should be $1 - 0.02 = 0.98$).
Correct model: $\boldsymbol{y = 500(0.98)^t}$

### Closing (Summary Example):
# Brief Explanations:
Today’s lesson covered exponential growth and decay. We learned to identify initial values, growth/decay factors, and formulate equations. We applied these to real - life scenarios like salary growth, stock decay, flu outbreaks, and population growth. Error analysis helped correct a wrong exponential model by understanding the decay factor.
# Answer:
Today’s lesson focused on exponential growth and decay. We learned to determine initial values, growth/decay factors, and create equations for real - world situations (e.g., salary, stock, flu, population). Error analysis was used to fix a wrong exponential model by correctly identifying the decay factor.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 3:

Step1: Identify Initial Value

Step2: Determine Growth Factor

Step3: Formulate the Equation

Step1: Identify Initial Value

Step2: Determine Growth Factor (Decay Factor)

Step3: Formulate the Equation

Answer:

Problem 4: