Questions: Life and Health Sciences 70. HIV Infection The time interval between a person's initial infection with HIV and that person's eventual development of AIDS symptoms is an important issue. The method of infection with HIV affects the time interval before AIDS develops. One study of HIV patients who were infected by intravenous drug use found that 17% of the patients had AIDS after 4 years, and 33% had developed the disease after 7 years. The relationship between the time interval and the percentage of patients with AIDS can be modeled accurately with a linear equation. Source: Epidemiologic Review. (a) Write a linear equation y=mt+b that models these data, using the ordered pairs (4,0.17) and (7,0.33). (b) Use your equation from part (a) to predict the number of years before half of these patients will have AIDS.

Life and Health Sciences
70. HIV Infection The time interval between a person's initial infection with HIV and that person's eventual development of AIDS symptoms is an important issue. The method of infection with HIV affects the time interval before AIDS develops. One study of HIV patients who were infected by intravenous drug use found that 17% of the patients had AIDS after 4 years, and 33% had developed the disease after 7 years. The relationship between the time interval and the percentage of patients with AIDS can be modeled accurately with a linear equation. Source: Epidemiologic Review.
(a) Write a linear equation y=mt+b that models these data, using the ordered pairs (4,0.17) and (7,0.33).
(b) Use your equation from part (a) to predict the number of years before half of these patients will have AIDS.
Transcript text: Life and Health Sciences 70. HIV Infection The time interval between a person's initial infection with HIV and that person's eventual development of AIDS symptoms is an important issue. The method of infection with HIV affects the time interval before AIDS develops. One study of HIV patients who were infected by intravenous drug use found that $17 \%$ of the patients had AIDS after 4 years, and $33 \%$ had developed the disease after 7 years. The relationship between the time interval and the percentage of patients with AIDS can be modeled accurately with a linear equation. Source: Epidemiologic Review. (a) Write a linear equation $y=m t+b$ that models these data, using the ordered pairs $(4,0.17)$ and $(7,0.33)$. (b) Use your equation from part (a) to predict the number of years before half of these patients will have AIDS.
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Solution

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Solution Steps

Solution Approach

(a) To write a linear equation \( y = mt + b \) that models the given data, we need to determine the slope \( m \) and the y-intercept \( b \). We can use the two given points \((4, 0.17)\) and \((7, 0.33)\) to find the slope \( m \) using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Once we have the slope, we can use one of the points to solve for \( b \).

(b) To predict the number of years before half of the patients will have AIDS, we set \( y = 0.5 \) in the linear equation derived in part (a) and solve for \( t \).

Step 1: Determine the Slope \( m \)

To find the slope \( m \) of the linear equation, we use the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Given the points \((4, 0.17)\) and \((7, 0.33)\): \[ m = \frac{0.33 - 0.17}{7 - 4} = \frac{0.16}{3} = 0.05333 \]

Step 2: Determine the Y-Intercept \( b \)

Using the slope \( m \) and one of the points, we can find the y-intercept \( b \) using the equation: \[ y = mt + b \] Substituting \( m = 0.05333 \) and the point \((4, 0.17)\): \[ 0.17 = 0.05333 \cdot 4 + b \] \[ 0.17 = 0.2133 + b \] \[ b = 0.17 - 0.2133 = -0.04333 \]

Step 3: Formulate the Linear Equation

The linear equation that models the data is: \[ y = 0.05333t - 0.04333 \]

Step 4: Predict the Number of Years Before Half of the Patients Have AIDS

To find the number of years \( t \) when half of the patients will have AIDS, set \( y = 0.5 \) and solve for \( t \): \[ 0.5 = 0.05333t - 0.04333 \] \[ 0.5 + 0.04333 = 0.05333t \] \[ 0.5433 = 0.05333t \] \[ t = \frac{0.5433}{0.05333} = 10.19 \]

Final Answer

\(\boxed{t = \frac{0.5 + 0.04333}{0.05333}} = \boxed{t = \frac{27}{5}} = \boxed{t = 5.4}\)

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