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110 000   -  mean   =

110 000 - 99 000   = 11 000       11 000/SD  =   11000/21000 = .5238 ~~ .52

             look at z = .52    on your  positive z - score table      z=.0..6950  or 69.5%

                                This is the cumulative area from the Left on the bell curve !!!

    you want the area to the RIGHT of this number   (GREATER than 110 000)    so   100% - 69.5% =  30.5% will have claims bigger than $110 000

Try #2 it is similar   and 3 is kinda similar

I do have a z score table but how is it going to help with the probability? I just started probability so I'm still a bit confused. Sorry.

You'll need your z-score tables.....do you know how to use them?

ALL of these questions require that you know     RATE  *  TIME = DISTANCE

Distance is the same for both ways on her trip      d

rate_u =x

rate_d =  x+22

 x * 6 =    d     on the way up to the mts       and        (x+22) * 4   =d    on the way back (they are equal d)

x*6   =   (x+22)*4    solve for x    her rate going up....then    

                                                               distance  =    rate * time  = x * 6

is it D ?

This is a standard linear programming problem, where we can apply the simplex method.

Applying the simplex method, we obtain these results:

0 0 1 3 5 16
0 0 2 1 8 20
0 9 -2 0 6 34
0 12 -8 -3 0 44
16 0 0 -7 -24 50
9 0 7 0 -3 800
8 0 8 1 0 1210
2 7 0 0 15 1300
4 6 0 -1 0 1470
0 60 10 0 1160

So, the minimum cost is $1160, which is obtained for 60 of Platter A and 10 of Platter B.

look at the peak of the triangle before it is translated.....

   it gets shifted +4 units in the x direction     and   + 5   in the y direction.....the point 5,2  gets similarly shifted,,,,you should be able to figure out what this final point would be....

I assume 'pre-image'   means the original image before the image is rotated or translated about the y or x axis   or shifted up/down or L/R

So the first one...rotate image clockwise  then shift right and down  and   then    H and C correspond so I think it is true.

Similarly on the second one     rotate clockwise 90 degrees  then shift right and down and  BA becomes  GI   so true.

Ha Ha!   I would not have thought of that ,,,,

rate   *     time  = distance

train time = t

bus time = b   =  31 hrs - t

80 * t         +      60 (31 - t)      =   2300

80t            +      1860  - 60 t     = 2300          solve for 't'   the train time

WHAT IS THE SECOND ONE

What is the smallest positive integer that has exactly 40 divisors?

\(\sigma_0(1680) = 40\)

I have updated the question

Let the number of adult tickets sold =A
Number of children's tickets sold      =1000 - A

2A + 1 * [1000 - A] =1850, solve for A
A = 850 adult tickets were sold.
1000 - 850 =150 children's tickets were sold

Let Joe's speed = S
Beatrice's speed =S + 2
Let his time        = T
Her time              = T - 1/2

S * T=12,   
(S+2) * (T - 1/2)=12, solve for S, T

S = 6 mph - Joe's speed
T = 2 hours - Joe's time
6 + 2 = 8 mph - Beatrice's speed
2 - 1/2 =1 1/2 hours - Beatrice's time

Looks like there are 6:

Edit:  Corrected.  Many thanks to heureka for alerting me to my error.

And you would be correct :)

You have edited this question twice but you do not appear to have proofread it.

1.   If WHAT is a root?

Good one !    

find the derivative using implicit differentiation. \(\tan(x - y) = \dfrac{y}{5 + x^2}\)

Derivative: \((5 + x^2)\tan(x - y) -y=0\)

Formula: \(y'(x) = -\dfrac{F_x}{F_y}\)

\(\small{ \begin{array}{|rcll|} \hline \mathbf{F_x = \ ?} \\ \hline F_x &=& \dfrac{d }{dx}(5 + x^2)\tan(x - y)- \dfrac{d}{dx}y \quad | \quad \dfrac{d}{dx}y = 0 \\ F_x &=& \dfrac{d }{dx}(5 + x^2)\tan(x - y)-0 \\ F_x &=& \dfrac{d }{dx}(5 + x^2)\tan(x - y) \\ F_x &=& (5 + x^2)\dfrac{d }{dx}\tan(x - y)+\tan(x - y)\dfrac{d }{dx}(5 + x^2) \quad | \quad \dfrac{d }{dx}(5 + x^2) = 2x \\ F_x &=& (5 + x^2)\dfrac{d }{dx}\tan(x - y)+2x\tan(x - y)\dfrac{d }{dx} \quad | \quad \dfrac{d }{dx}\tan(x - y) = \dfrac{1}{\cos^2(x-y)}*1 \\ F_x &=& (5 + x^2)\dfrac{1}{\cos^2(x-y)}*1+2x\tan(x - y) \\ F_x &=& \dfrac{5 + x^2+2x\tan(x - y)\cos^2(x-y)}{\cos^2(x-y)} \\ F_x &=& \dfrac{5 + x^2+2x\sin(x - y)\cos(x-y)}{\cos^2(x-y)} \quad | \quad 2\sin(x - y)\cos(x-y) = \sin\Big(2(x-y)\Big) \\ \mathbf{F_x} &=& \mathbf{\dfrac{5 + x^2+x\sin\Big(2(x-y)\Big)}{\cos^2(x-y)}} \\ \hline \end{array} }\)

\(\small{ \begin{array}{|rcll|} \hline \mathbf{F_y = \ ?} \\ \hline F_y &=& \dfrac{d }{dy}(5 + x^2)\tan(x - y)- \dfrac{d}{dy}y \quad | \quad \dfrac{d}{dy}y = 1 \\ F_y &=& \dfrac{d }{dy}(5 + x^2)\tan(x - y)-1 \\ F_y &=& (5 + x^2)\dfrac{d }{dy}\tan(x - y)-1 \quad | \quad \dfrac{d }{dx}\tan(x - y) = \dfrac{1}{\cos^2(x-y)}*(-1) \\ F_y &=& (5 + x^2)\dfrac{1}{\cos^2(x-y)}*(-1) -1 \\ F_y &=& -\left( \dfrac{(5 + x^2)}{\cos^2(x-y)} +1 \right) \\ F_y &=& -\left( \dfrac{5 + x^2+\cos^2(x-y)}{\cos^2(x-y)} \right) \quad | \quad \cos^2(x-y)=\dfrac{1+ \cos\Big(2(x-y)\Big)}{2} \\ F_y &=& -\left( \dfrac{5 + x^2+\dfrac{1+ \cos\Big(2(x-y)\Big)}{2}}{\cos^2(x-y)} \right) \\ F_y &=& -\left( \dfrac{10 + 2x^2+1+ \cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} \right) \\ \mathbf{F_y} &=& \mathbf{ -\left( \dfrac{11 + 2x^2+\cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} \right)} \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline y'(x) &=& -\dfrac{F_x}{F_y} \\\\ y'(x) &=& -\dfrac{\dfrac{5 + x^2+x\sin\Big(2(x-y)\Big)}{\cos^2(x-y)}} {-\left( \dfrac{11 + 2x^2+\cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} \right)} \\\\ y'(x) &=& \dfrac{\dfrac{5 + x^2+x\sin\Big(2(x-y)\Big)}{\cos^2(x-y)}} {\dfrac{11 + 2x^2+\cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} } \\\\ y'(x) &=& \dfrac{ 2\left( 5 + x^2+ x\sin \Big(2(x-y)\Big) \right) } { 11 + 2x^2 +\cos\Big(2(x-y)\Big)} \\\\ \mathbf{y'(x)} &=& \mathbf{\dfrac{ 2\Big( 5 + x^2+ x\sin(2x-2y) \Big) } { 11 + 2x^2 +\cos(2x-2y)} } \\ \hline \end{array}\)

The sum of the lengths of the edges of the pyramid is    6*6 + 3*6 = 54 

ABCD is a isosceles trapezoid with AB=10, DB=8 and angle A=120 degrees what is the length of CD

( This question is not written correctly. Cannot be DB=8 It's either CB=8, or DA=8)

Angle FBC = 120° - 90° = 30°

FC = sin(∠FBC) * 8  = 4

CD = AB + 2 (FC) = 18 units  

Find the derivative of the following via implicit differentiation:
d/dx(tan(x - y)) = d/dx(y/(5 + x^2))

Using the chain rule, d/dx(tan(x - y)) = ( dtan(u))/( du) ( du)/( dx), where u = x - y and d/( du)(tan(u)) = sec^2(u):
(d/dx(x - y)) sec^2(x - y) = d/dx(y/(5 + x^2))

Differentiate the sum term by term and factor out constants:
d/dx(x) - d/dx(y) sec^2(x - y) = d/dx(y/(5 + x^2))

The derivative of x is 1:
sec^2(x - y) (-(d/dx(y)) + 1) = d/dx(y/(5 + x^2))

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
sec^2(x - y) (1 - (d/dx(x)) y'(x)) = d/dx(y/(5 + x^2))

The derivative of x is 1:
sec^2(x - y) (1 - 1 y'(x)) = d/dx(y/(5 + x^2))

Use the quotient rule, d/dx(u/v) = (v ( du)/( dx) - u ( dv)/( dx))/v^2, where u = y and v = x^2 + 5:
sec^2(x - y) (1 - y'(x)) = ((5 + x^2) (d/dx(y)) - y (d/dx(5 + x^2)))/(5 + x^2)^2

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
sec^2(x - y) (1 - y'(x)) = ((5 + x^2) (d/dx(x)) y'(x) - (d/dx(5 + x^2)) y)/(5 + x^2)^2

The derivative of x is 1:
sec^2(x - y) (1 - y'(x)) = (-(d/dx(5 + x^2)) y + 1 (5 + x^2) y'(x))/(5 + x^2)^2

Differentiate the sum term by term:
sec^2(x - y) (1 - y'(x)) = (-y d/dx(5) + d/dx(x^2) + (5 + x^2) y'(x))/(5 + x^2)^2

The derivative of 5 is zero:
sec^2(x - y) (1 - y'(x)) = (-(d/dx(x^2) + 0) y + (5 + x^2) y'(x))/(5 + x^2)^2

Simplify the expression:
sec^2(x - y) (1 - y'(x)) = (-(d/dx(x^2)) y + (5 + x^2) y'(x))/(5 + x^2)^2

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2.
d/dx(x^2) = 2 x:
sec^2(x - y) (1 - y'(x)) = (-y 2 x + (5 + x^2) y'(x))/(5 + x^2)^2

Expand the left hand side:
sec^2(x - y) - sec^2(x - y) y'(x) = (-2 x y + (5 + x^2) y'(x))/(5 + x^2)^2

Expand the right hand side:
sec^2(x - y) - sec^2(x - y) y'(x) = -(2 x y)/(5 + x^2)^2 + (5 y'(x))/(5 + x^2)^2 + (x^2 y'(x))/(5 + x^2)^2

Subtract (x^2 y'(x))/(x^2 + 5)^2 + (5 y'(x))/(x^2 + 5)^2 from both sides:
sec^2(x - y) - (5 y'(x))/(5 + x^2)^2 - (x^2 y'(x))/(5 + x^2)^2 - sec^2(x - y) y'(x) = -(2 x y)/(5 + x^2)^2

Subtract sec^2(x - y) from both sides:
-(5 y'(x))/(5 + x^2)^2 - (x^2 y'(x))/(5 + x^2)^2 - sec^2(x - y) y'(x) = -sec^2(x - y) - (2 x y)/(5 + x^2)^2

Collect the left hand side in terms of y'(x):
(-5/(5 + x^2)^2 - x^2/(5 + x^2)^2 - sec^2(x - y)) y'(x) = -sec^2(x - y) - (2 x y)/(5 + x^2)^2

Divide both sides by -x^2/(x^2 + 5)^2 - 5/(x^2 + 5)^2 - sec^2(x - y):

y'(x) = (-sec^2(x - y) - (2 x y)/(5 + x^2)^2)/(-5/(5 + x^2)^2 - x^2/(5 + x^2)^2 - sec^2(x - y))

Courtesy of Mathematica 11 Home Edition.

like this F(IV)E.....

What is the largest seven-digit number divisible by 44 that can be formed by the digits 1, 2, 3, 4, 6, 7, and 8 each used exactly once?

Well it must be divisible by 4 which means the last two digits must be divisible by 4

So it could end in 12, 16, 24, 28, 32, 36, 48, 64, 68, 72, 76, 84,

And

It must be divisible by 11