Consider removing the point (33,-61) and calculating a new least-squares regression line

"i will have the report ready for you review by the end of the day." it is quick and to the point we have an appointment at 3: 00. emphasized great to see how this person is feeling

Yo idk the answer so here i am

Iwould say b but i’m not so sure

You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule.    set the derivative to equal -1 and solve for x.      take log base 10 of both sides. also recall that  .    so the x-coord of the point of tangency is .the y-coordinate of this point of tangency can be calculated using the original function:     using point-slope form, the equation of the tangent line is      the x-intercept of this tangent line is when the graph is at y=0.    this rounds off to  .choice c.