Pls help I need in 1 hour this is all data The data show the bug chirpe per minute at difforent termperaturee. Find the regression equation, letting the firet variable be the independent (x) variable. Find the beet predictod tomperature for a time when a bug ia chirping et the min e3000 chirpe per minute. Use a aignifioance level of 0.05. What is wrong with this predicted value?Chirpe In 1 minTemperature ("F)/1170 1223 1129 1145 1020 824|92.4 86.3 89.4 85.8 81.3 714What is the regreesion equation?y-/0.002) +0. 134x(Round the x-coefficiont to four decimal places as neoded. Round the constant to two decimal places as needed.)What is the best predicted temperature for a time when a bug is chirping at the rate of 3000 chirpa per minute?The best predicted tomperature when a bug is chirping at 3000 chirps per minuto is (00.05 °F(Round to one deaimal place as needed.)What is wrong with this predicted value? Choose the correct answer below.2 A. The first variable should have been the dependent variable.B. It is unnrealistically high. The value 3000 is far outside of the range of observed valuos.C. H is only an approximation. An unrounded value would be considered accurate.Nothingwrong with this value. It can be troated as an accurate prediction.

Given data: x y 1170 92.4 1223 86.3 1129 89.4 1145 85.8 1029 81.9 824 71.4The table for necessary calculation: x y x-x¯ y-y¯ x-x¯2 y-y¯2 x-x¯y-y¯ 1 1170 92.4 83.333 7.8667 6944.389 61.88497 655.5557 2 1223 86.3 136.333 1.7667 18586.69 3.121229 240.8595 3 1129 89.4 42.333 4.8667 1792.083 23.68477 206.022 4 1145 85.8 58.333 1.2667 3402.739 1.604529 73.89041 5 1029 81.9 -57.667 -2.6333 3325.483 6.934269 151.8545 6 824 71.4 -262.667 -13.1333 68993.95 172.4836 3449.685 Sum 6520 507.2 103045.3 269.7133 4777.867 Average 1086.667 84.5333 The regression line is y=a+bx. b=∑x-x¯y-y¯∑x-x¯2 =4777.867103045.3 ≈0.0464 a=y¯-bx¯ =84.5333-4777.867103045.3×1086.667 ≈34.15 y^=34.15+0.0464x Obtain the best-predicted temperature for a time when a bug chirps at the rate of 3000 chirps per minute. Sub x=3000 in the regression equation. y^x=3000=34.15+0.0464×3000 =173.35 Thus, the best predicted temperature for a time when a bug chirping at the rate of 3000 chirps per minute is 173.35 F. Determine what is wrong with this predicted temperature. It is clear that the chirps 3000 times in 1 minute lie outside the range of observed values it can be concluded that the obtained predicted temperature would be unrealistically high. Correct option: Option B