(Enter only a number in the response box, not the units)    The yield of wool in a particular farm is modeled by a linear function  P(t)=90t+3,450P𝑡=90𝑡+3,450 , where  P(t)𝑃𝑡  is measured in kg and  t𝑡  in years.     What is the initial production?     P(0)=𝑃0=    kg.     What is the prediction for the production in 8 years? P(8)=𝑃8=     kg.         The value in 8 years is   P(8)=4,056𝑃8=4,056  kg.     What is the absolute error of this model at  t=8𝑡=8 ?  Absolute error =     kg.     What is the percentage error? Round the answer to one decimal place.               Percentage error =     %

Consider the following output for a linear model for estimating the salary of employees at a company (in $) by the number of years employed at the company. > technitron.lm = lm(salary ~ yrs.empl, data = technitron.df)> newdata = list(yrs.empl = c(6, 8))> predict(technitron.lm, newdata, interval = "confidence") fit lwr upr1 35037.47 32582.25 37492.692 37251.91 35024.31 39479.50> predict(technitron.lm, newdata, interval = "prediction") fit lwr upr1 35037.47 20396.88 49678.062 37251.91 22647.76 51856.05Give the appropriate interval for the average salary of employees who have been employed for 6 years.Group of answer choices(32582.25, 37492.69)(20396.88, 49678.06)(35024.31, 39479.50)(22647.76, 51856.05)

A farm produced 1,626 kg wool in year 2,015 and 1,589 kg wool in year 2,019. What was the rate of change of the wool production?

PLANT X= A:1.0 B: 1.5 C: 2.0 D: 2.5 E: 3.0 F: 3.5 G: 4 H: 4.5 PLANT Y= A:3.9 B: 4.4 C: 5.8 D: 6.6 E: 7.0 F: 7.1 G: 7.3 H: 7.7 a) Plot a scatter diagram of yield y, against amount of fertilizer x. (b) calculate the equation of the least squares regression line of y on x (c) estimate the yield of a plant treated weekly with 3.2grams of fertilizer. (d) explain why it may not be appropriate t use your equation to predict the yield of a tomato plant treated weekly with 20grams of fertilizer. QUESTION 3 Students were given different drug treatment before revising for their examinations. some were given a memory drug, some a placebo and others no treatment. the exam scores (%) are shown for the three groups: MEMORY DRUG: 70,77,83,90,97. Mean: 83.40, variance: 112.40 PLACEBO: 37,43,50, 57, 63. mean:50.00, variance: 109.00 NO TREATMENT: 2, 10, 17, 23, 30 . Mean: 16.60, Variance: 112.30 GRAND MEAN: 50.00 GRAND VARIANCE: 892.14 (A) Carry out an analysis of variance to determine wether the treatments have different effects on the performance of the student (use alpha = 0.05)

predicted received = 6.922 + 1.053 *sent，If, on a particular day, the lecturer sends 40 emails and receives 40 emails, that would mean that the prediction error for this day is approximately 9. Using the training data of 159 emails and the equation of the fitted line in Question 11, determine an appropriate error amount for a prediction model. Write the equation for this prediction model which includes the error term. In no more than one sentence, discuss a possible limitation of using this model to calculate a prediction interval for the number of emails received by this lecturer on a day where the lecturer sends 80 emails.

Suppose for a particular week, the forecasted sales were $4,000. The actual sales were $3,000.  What is the value of the mean absolute percentage error?