TutorMe homepage
Subjects
PRICING
COURSES
SIGN IN
Start Free Trial
Kirsten M.
Research Analyst
Tutor Satisfaction Guarantee
Calculus
TutorMe
Question:

Differentiate $$y=\sqrt{13x^{2} - 5x +8}$$

Kirsten M.
Answer:

$$y' =(1/2)(13x^{2} - 5x +8) ^ {-1/2} (26x-5)$$ $$y' =\frac{26x-5}{ 2 \sqrt{13x^{2} - 5x +8}}$$

Statistics
TutorMe
Question:

Two dice are rolled. A = ‘sum of two dice equals 3’ B = ‘sum of two dice equals 7’ C = ‘at least one of the dice shows a 1’ (a) What is P(A|C)? (b) What is P(B|C)? (c) Are A and C independent? What about B and C?

Kirsten M.
Answer:

Each outcome is equally likely, with probability 1/36 A = {(1, 2), (2, 1)}, B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)} (a) $$P(A|C) = \frac{P(A ∩ C)}{P(C)} = \frac{2/36}{11/36} = \frac{2}{11 }$$ (b) $$P(B|C) = \frac{P(B ∩ C)}{P(C)} = \frac{2/36}{11/36} = \frac{2}{11}$$ (c) $$P(A) = \frac{2}{36} \neq P(A|C)$$, so they are not independent. Similary, $$P(B) = \frac{6}{36} \neq P(B|C)$$, so they are no independent

Algebra
TutorMe
Question:

Given the line 2x – 3y = 9 and the point (4, –1), find lines, in slope-intercept form, through the given point such that the two lines are, respectively,: (a) parallel to the given line, and (b) perpendicular to it

Kirsten M.
Answer:

We start by recognizing the equation 2x -3y =9 is in the standard form. To make our work easier, and for preference, we transform the equation into slope intercept form (reference the equations below) Standard Form of an Equation Ax+By=C Slope Intercept Form y=mx+b where: m=slope b= y-intercept Point Slope Form y-y1=m(x-x1) where: y1= y-corrdinate x1= x-corrdinate (corresponding to y-coordinate) m= slope Slope Intercept Form of Given Equation : 2x-3y=9 -3y=9-2x (-3y)/(-3)=(9-2x)/(-3) y= -3 + (2/3)x y= (2/3)x-3 m=2/3 b= -3 Part (A) For two lines to be parallel to each other, they must have the same slope and we are given that the line must go through the point (4, -1). Above we found the slope intercept form of the given equation and found our slope value (m-value) to be 2/3. We now plug these value into the point slope of the equation. y- (-1)= (2/3)(x-4) y+1=(2/3)x -(8/3) y=(2/3)x - (11/3) ANSWER Part (B) The slope of a line perpendicular to the given equation would be negative reciprocal of the given lines slope. For example we have: m = 2/3 Reciprocal m= 3/2 Negative Reciprocal m=-3/2 We no simply use this new slope and the given points (4, -1) and plug them into the point slope equation and solve for y to ensure the equation is in slope intercept form. y- (-1)= (-3/2)(x-4) y+1=(-3/2)x+6 y= (-3/2)x+5 ANSWER

Send a message explaining your
needs and Kirsten will reply soon.
Contact Kirsten
Ready now? Request a lesson.
Start Session
FAQs
What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Session" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.