b) use the quadratic polynomial to estimate the outdoor temperature at 9:30 am, to the nearest tenth of a degree c) use the quadratic polynomial y=-0.35t2+11t-6 together with algebra to estimate the times of day when the outdoor temperature y was 70 degrees. Solve the quadratic equation y=70=-0.35t2+11t-6.
b) use the quadratic polynomial to estimate the outdoor temperature at 9:30 am, to the nearest tenth of a degree c) use the quadratic polynomial y=-0.35t2+11t-6 together with algebra to estimate the times of day when the outdoor temperature y was 70 degrees. Solve the quadratic equation y=70=-0.35t2+11t-6.
b) use the quadratic polynomial to estimate the outdoor temperature at 9:30 am, to the nearest tenth of a degree c) use the quadratic polynomial y=-0.35t2+11t-6 together with algebra to estimate the times of day when the outdoor temperature y was 70 degrees. Solve the quadratic equation y=70=-0.35t2+11t-6.
b) use the quadratic polynomial to estimate the outdoor temperature at 9:30 am, to the nearest tenth of a degree
c) use the quadratic polynomial y=-0.35t2+11t-6 together with algebra to estimate the times of day when the outdoor temperature y was 70 degrees. Solve the quadratic equation y=70=-0.35t2+11t-6. Report the times to the nearest quarter hour.
Transcribed Image Text:**Data:**
On a particular summer day, the outdoor temperature was recorded at 8 times of the day. The following table was compiled. A scatterplot was produced, and the quadratic (parabola) of best fit was determined.
| t = Time of day (hour) | y = Outdoor Temperature (degrees F) |
|------------------------|------------------------------------|
| 7 | 52 |
| 9 | 67 |
| 11 | 73 |
| 13 | 76 |
| 14 | 78 |
| 17 | 79 |
| 20 | 76 |
| 23 | 61 |
**Graph Description:**
The graph titled "Temperature on a Summer Day" plots the temperature (in degrees Fahrenheit) on the vertical axis against the time of day (in hours, since midnight) on the horizontal axis. The plotted points follow a parabolic trend, fitted by the quadratic equation \( y = -0.35t^2 + 11t - 6 \) with a coefficient of determination \( R^2 = 0.97 \), indicating a strong fit.
**Quadratic Polynomial of Best Fit:**
\[ y = -0.35t^2 + 11t - 6 \]
where \( t = \) Time of day (hour) and \( y = \) Temperature (in degrees F).
**Remarks:**
- The times are the hours since midnight. For example, \( t = 7 \) means 7 am, and 13 means 1 pm.
**Instructions:**
(a) Using algebraic techniques that have been learned, find the maximum temperature predicted by the quadratic model and determine the time when it occurred.
- Report the time to the nearest quarter hour (i.e., :00, :15, :30, or :45). For example, a time of 18.25 hours is reported as 6:15 pm.
- Report the maximum temperature to the nearest tenth of a degree.
- Show algebraic work.
**Conversion Table (Decimal to Minutes):**
| Decimal | Minute |
|---------|--------|
| 0.00 | :00 |
| 0.25 | :15 |
| 0.50 | :30 |
| 0.75 | :45 |
**Note:**
*Be
Formula Formula A polynomial with degree 2 is called a quadratic polynomial. A quadratic equation can be simplified to the standard form: ax² + bx + c = 0 Where, a ≠ 0. A, b, c are coefficients. c is also called "constant". 'x' is the unknown quantity
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![**Data:**
On a particular summer day, the outdoor temperature was recorded at 8 times of the day. The following table was compiled. A scatterplot was produced, and the quadratic (parabola) of best fit was determined.
| t = Time of day (hour) | y = Outdoor Temperature (degrees F) |
|------------------------|------------------------------------|
| 7 | 52 |
| 9 | 67 |
| 11 | 73 |
| 13 | 76 |
| 14 | 78 |
| 17 | 79 |
| 20 | 76 |
| 23 | 61 |
**Graph Description:**
The graph titled "Temperature on a Summer Day" plots the temperature (in degrees Fahrenheit) on the vertical axis against the time of day (in hours, since midnight) on the horizontal axis. The plotted points follow a parabolic trend, fitted by the quadratic equation \( y = -0.35t^2 + 11t - 6 \) with a coefficient of determination \( R^2 = 0.97 \), indicating a strong fit.
**Quadratic Polynomial of Best Fit:**
\[ y = -0.35t^2 + 11t - 6 \]
where \( t = \) Time of day (hour) and \( y = \) Temperature (in degrees F).
**Remarks:**
- The times are the hours since midnight. For example, \( t = 7 \) means 7 am, and 13 means 1 pm.
**Instructions:**
(a) Using algebraic techniques that have been learned, find the maximum temperature predicted by the quadratic model and determine the time when it occurred.
- Report the time to the nearest quarter hour (i.e., :00, :15, :30, or :45). For example, a time of 18.25 hours is reported as 6:15 pm.
- Report the maximum temperature to the nearest tenth of a degree.
- Show algebraic work.
**Conversion Table (Decimal to Minutes):**
| Decimal | Minute |
|---------|--------|
| 0.00 | :00 |
| 0.25 | :15 |
| 0.50 | :30 |
| 0.75 | :45 |
**Note:**
*Be](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39e26e0d-5044-4844-9595-6ca9cde14cdd%2F0b6eb67a-bc07-4daf-adf7-819475e7702c%2Fp5w8min_processed.jpeg&w=3840&q=75)
