The best predicted temperature at a time when a cricket chirps 3000 times in 1 minute, based on the regression equation, is ______________ Fahrenheit. (Round to the nearest integer as needed.) What is wrong with this predicted temperature? A. It is unrealistically high. The value 3000 is far outside of the range of observed values. B. It is only an approximation. An unrounded value would be considered accurate. C. The chirps in 1 minute should have been the dependent variable. D.Nothing is wrong with this value. It can be treated as an accurate prediction. **Educational Website Content: Analyzing Cricket Chirps and Temperature**### Problem StatementThe table below lists the number of cricket chirps in one minute and the corresponding temperatures in degrees Fahrenheit (°F). Your task is to find the regression equation, using chirps in one minute as the independent variable \((x)\). Then, predict the temperature when cricket chirps occur 3000 times in one minute using the regression equation. Discuss any potential issues with this predicted temperature.| Chirps in 1 min | 791 | 980 | 1008 | 1211 | 904 | 1032 | 1070 | 961 ||-----------------|-----|-----|------|------|-----|------|------|-----|| Temperature (°F) | 72.1 | 80.1 | 82.5 | 91.8 | 79.8 | 83.8 | 86.7 | 74.5 |### Task1. **Find the Regression Equation:** - The regression equation is represented as \(\hat{y} = a + bx\), where \(\hat{y}\) is the predicted temperature. - Round the y-intercept \(a\) to one decimal place. - Round the slope \(b\) to four decimal places.2. **Prediction and Analysis:** - Use the regression equation to predict the temperature when there are 3000 chirps in one minute. - Evaluate any issues with the prediction.### Detailed Explanation- The regression equation is vital for establishing the relationship between two variables: cricket chirps and temperature.- Predicting temperatures far outside the observed data range, such as 3000 chirps per minute, may lead to inaccurate predictions due to extrapolation beyond the data's range.

Given: n = 8 Formula Used: The regression equation is: Y = a + bX b = n∑XY-∑X∑Yn∑X2-∑X2 a = ∑Y-b∑XnExplanation: Chirps in 1 min X Temperature Y X2 XY 791 72.1 625681 57031.1 980 80.1 960400 78498 1008 82.5 1016064 83160 1211 91.8 1466521 111169.8 904 79.8 817216 72139.2 1032 83.8 1065024 86481.6 1070 86.7 1144900 92769 961 74.5 923521 71594.5 ∑X = 7957 ∑Y = 651.3 ∑X2 = 8019327 ∑XY = 652843.2 b = n∑XY-∑X∑Yn∑X2-∑X2= 8×652843.2-7957×651.38×8019327-63313849= 0.0480 a = ∑Y-b∑Xn= 651.3-0.048×79578= 33.7 The regression equation is: Y = 33.7 + 0.0480XThe regression equation is: Y = 33.7 + 0.0480X Substitute the value of X = 3000 in the above equation. Y = 33.7 + 0.0480XY = 33.7 + 0.0480×3000Y = 177.7Y ≈ 178 {Rounded to nearest integer} B. It is only an approximation. An unrounded value would be considered accurate.Answer: The regression equation is: Y = 33.7 + 0.048X The best-predicted temperature at a time when a cricket chirps 3000 times in 1 minute, based on the regression equation, is 178 Fahrenheit. B). It is only an approximation. An unrounded value would be considered accurate. Option (B) is correct.