a) Find the mean, median, and mode of $163, 149, 155, 162, 168, 168$. b) Solve the equation $2^{2x} - 9(2^x) + 8 = 0$. c) Express $3 \sin \theta + 5 \cos \theta$ in the form $R \sin (\theta + \alpha)$, and hence solve the equation $3 \sin \theta + 5 \cos \theta = 4$, for values of $\theta$ between $0^{\circ}$ and $360^{\circ}$.
Answer (2)
The mean of the dataset is approximately 160.83, the median is 162.5, and the mode is 168. The solutions to the equation are x = 0 and x = 3. The values of theta that solve the equation are approximately 77.65° and 344.27°.
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Calculate the mean of the data set: mean = 6 965 ≈ 160.83 .
Find the median of the data set: median = 2 162 + 163 = 162.5 .
Identify the mode of the data set: mode = 168 .
Solve the exponential equation 2 2 x − 9 ( 2 x ) + 8 = 0 , obtaining solutions x = 0 and x = 3 .
Express 3 sin θ + 5 cos θ as 34 sin ( θ + 59.0 4 ∘ ) .
Solve 3 sin θ + 5 cos θ = 4 , finding θ ≈ 77.6 5 ∘ and θ ≈ 344.2 7 ∘ .
The solutions for θ are 77.6 5 ∘ , 344.2 7 ∘ .
Explanation
Understanding the Data We are given the data set 163 , 149 , 155 , 162 , 168 , 168 and asked to find the mean, median, and mode.
Calculating the Mean The mean is the average of the numbers. We sum the numbers and divide by the count: mean = 6 163 + 149 + 155 + 162 + 168 + 168 = 6 965 = 160.8333... Rounding to two decimal places, the mean is approximately 160.83 .
Finding the Median To find the median, we first sort the data in ascending order: 149 , 155 , 162 , 163 , 168 , 168 . Since there are an even number of data points, the median is the average of the two middle numbers: median = 2 162 + 163 = 2 325 = 162.5
Identifying the Mode The mode is the number that appears most frequently. In this data set, 168 appears twice, which is more than any other number. Therefore, the mode is 168 .
Transforming the Equation We are given the equation 2 2 x − 9 ( 2 x ) + 8 = 0 and asked to solve for x . Let y = 2 x . Then the equation becomes: y 2 − 9 y + 8 = 0
Solving the Quadratic Equation We can factor the quadratic equation as follows: ( y − 1 ) ( y − 8 ) = 0 So, y = 1 or y = 8 .
Finding the Values of x Substituting back y = 2 x , we have 2 x = 1 or 2 x = 8 .
If 2 x = 1 , then x = 0 since 2 0 = 1 .
If 2 x = 8 , then x = 3 since 2 3 = 8 .
Thus, the solutions are x = 0 and x = 3 .
Expressing in the Required Form We are asked to express 3 sin θ + 5 cos θ in the form R sin ( θ + α ) and solve the equation 3 sin θ + 5 cos θ = 4 for 0 ∘ < θ < 36 0 ∘ .
Using the identity R sin ( θ + α ) = R ( sin θ cos α + cos θ sin α ) = ( R cos α ) sin θ + ( R sin α ) cos θ , we can equate coefficients: R cos α = 3 and R sin α = 5
Calculating R and Alpha To find R , we use the Pythagorean identity: R 2 = ( R cos α ) 2 + ( R sin α ) 2 = 3 2 + 5 2 = 9 + 25 = 34 So, R = 34 ≈ 5.83 .
To find α , we have: tan α = R cos α R sin α = 3 5 α = arctan ( 3 5 ) ≈ 59.0 4 ∘
Solving the Equation Now we solve the equation 3 sin θ + 5 cos θ = 4 , which can be written as: 34 sin ( θ + 59.0 4 ∘ ) = 4 sin ( θ + 59.0 4 ∘ ) = 34 4 ≈ 0.686
Finding Beta Let β = θ + 59.0 4 ∘ . Then sin β = 0.686 .
The principal value of β is: β 1 = arcsin ( 0.686 ) ≈ 43.3 1 ∘ The other solution in the range 0 ∘ to 36 0 ∘ is: β 2 = 18 0 ∘ − 43.3 1 ∘ = 136.6 9 ∘
Finding Theta Now we solve for θ :
θ 1 = 43.3 1 ∘ − 59.0 4 ∘ = − 15.7 3 ∘ Since we want values between 0 ∘ and 36 0 ∘ , we add 36 0 ∘ to get θ 1 = 344.2 7 ∘ .
θ 2 = 136.6 9 ∘ − 59.0 4 ∘ = 77.6 5 ∘
Final Solutions Therefore, the solutions for θ are approximately 77.6 5 ∘ and 344.2 7 ∘ .
Examples
Understanding mean, median, and mode is crucial in analyzing data sets, such as test scores or sales figures. Solving exponential equations is essential in modeling growth and decay processes, like population dynamics or radioactive decay. Expressing trigonometric functions in different forms helps in analyzing oscillatory phenomena, such as sound waves or alternating current circuits. These mathematical tools provide valuable insights in various real-world applications.
