Can anyone help me with this??

Mathematics

2 answers:

kotegsom [21]2 years ago

8 0

Well calculate the win rate of both tracks

Dry track
16/64 = 0.25 = 25%

Wet track
12/40 = 0.3 = 30%

Dave was more successful on the wet track

Rudiy272 years ago

4 0

I think that’s the answer

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Find the tangent plane to the given surface of f(x,y)=6- 6/5 x-y at the point (5, -1, 1). Make sure tat your final answer for th

HACTEHA [7]

Answer:

Required equation of tangent plane is $z=\frac{6}{5}(x-5y-11)$ .

Step-by-step explanation:

Given surface function is,

$f(x,y)=6-\frac{6}{5}(x-y)$

To find tangent plane at the point (5,-1,1).

We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)$

So that,

$f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}$

$f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}$

$f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6$

Substitute all these values in (1) we get,

$z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}$

$\therefore z=\frac{6}{5}(x-5y-11)$

Which is the required euation of tangent plane.

4 0

2 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

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