# Gym Membership Cost Comparison Let's solve each part step-by-step. --- ## a) Linear Function for Gym A Gym A: - Initial sign-up fee: \$40 - Monthly fee: \$25 Let \( x \) be the number of months. **Linear function:** \[ C_A(x) = 40 + 25x \] --- ## b) Linear Function for Gym B Gym B: - Initial sign-up fee: \$10 - Monthly fee: \$30 **Linear function:** \[ C_B(x) = 10 + 30x \] --- ## c) Months When Costs Are the Same Set the two cost functions equal: \[ C_A(x) = C_B(x) \] \[ 40 + 25x = 10 + 30x \] Solve for \( x \): 1. Subtract \( 25x \) from both sides: \[ 40 = 10 + 5x \] 2. Subtract 10 from both sides: \[ 30 = 5x \] 3. Divide both sides by 5: \[ x = 6 \] **The costs are equal after 6 months.** --- ## d) What Is That Cost? Substitute \( x = 6 \) into either cost function: \[ C_A(6) = 40 + 25 \times 6 = 40 + 150 = 190 \] \[ C_B(6) = 10 + 30 \times 6 = 10 + 180 = 190 \] **The cost is \$190.** --- ## Summary Table | Gym | Cost Function | Cost at 6 Months (\$) | |-------|----------------------|----------------------| | A | \( C_A(x) = 40+25x \) | 190 | | B | \( C_B(x) = 10+30x \) | 190 | --- ## Final Answers - **a)** \( C_A(x) = 40 + 25x \) - **b)** \( C_B(x) = 10 + 30x \) - **c)** Costs are equal after **6 months** - **d)** The equal cost is **\$190**

# Probability of Waiting Time for Exponentially Distributed Calls Given: - Mean waiting time \( \mu = 2 \) hours - Rate \( \lambda = \frac{1}{\mu} = \frac{1}{2} \) calls per hour ## Exponential Distribution The probability density function (PDF) for an exponentially distributed random variable \( X \) is given by: \[ f(x) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0 \] The cumulative distribution function (CDF) is: \[ F(x) = 1 - e^{-\lambda x} \] ## Problem Statement We want to find the probability that no call arrives in the first 3 hours, and the next call arrives after an additional 2 hours (i.e., after 5 hours total). ### Step 1: No Call in the First 3 Hours The probability that no call arrives in the first 3 hours is given by: \[ P(X > 3) = 1 - F(3) = e^{-\lambda \cdot 3} = e^{-\frac{1}{2} \cdot 3} = e^{-1.5} \] ### Step 2: Next Call After Additional 2 Hours Given that no calls have arrived in the first 3 hours, we need the probability that the next call arrives after an additional 2 hours. For an exponential distribution, the memoryless property tells us that: \[ P(X > s + t | X > s) = P(X > t) \] Thus, we can rewrite our probability: \[ P(X > 2 | X > 3) = P(X > 2) = e^{-\lambda \cdot 2} = e^{-\frac{1}{2} \cdot 2} = e^{-1} \] ### Step 3: Combined Probability The overall probability is the product of the two independent events: \[ P(X > 3) \cdot P(X > 2) = e^{-1.5} \cdot e^{-1} = e^{-2.5} \] ## Final Answer The probability that the next call arrives after an additional 2 hours, given that no call arrived in the first 3 hours, is: \[ P = e^{-2.5} \approx 0.0821 \] Thus, the final answer is approximately **0.0821** or **8.21%**.