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well your answer agrees with sim!

what's the extra factor of 4! for?

oh nm, it's because you are having order matter.

(sorry I edited accidentally, I think it is back to how it was now)

Hi Rom,

I worked it out before and got an answer of 0.14

I have re-shown my working now.

I thought it was wrong because your logic seemed better but maybe it is not wrong afterall.

Above is wrong.  Simulation is showing p is approximately 0.14

Still working on this.

im sorry to trouble you i have one more question.

im not sure how to set this up 

"Write the equation of the elliptical orbit of Mercury, where the major axis runs horizontally. Allow a and b to be measured in millions of miles. Use the origin as the center of the ellipse."

Picture for number one?

2)

Just solve the equation normally, but show EACH STEP of your work.

Now take each step of your work and justify each step.

You should use different properties (such as additive property of equality for example) that you learned in class..

I dont this is really hard but I have a crazy suggestion

Draw a regular hexagon inscribed in the circle

Now notice how the radii meet at the center.

Adjust that regular center so it fits the center of your problem.

Now imagine the radii adjusting as well.

Does that show that it is concurrent?

Welcome!

thank you both so much!!

Can someone please help? The due date is Friday and I am really stuck, thank you for any help!

2)

We can try to find the distance of the perihelion to the center of the ellipse.

116 / 2 = 58 million miles

58 - 46 (distance from perhelion to sun) = 12 million miles between sun and center of ellipse

Not sure about my answers never learned calculus

This is basically what the question is asking

Lets say the sun is located at the focus of the left side of the image

____________________________________________________

So the distance between the perihelion and aphelion should be 46 + 70 = 116 million miles??

I haven't learned much in calculus yet this is as much as I can give you

Thanks melody!

help me expand this calculation so that I understand the result please. why is 1-(10/75) = 13/15 ?    

CU's way is very good, but you could reduce 10/75 first if you wanted to. 

we're starting with                                                               1 – (10/75) 

reduce 10/75 by dividing both top and bottom by 5            1 – (2/15)

since 1 = 15/15                                                                   (15/15) – (2/15)  =  13/15 

.

Rom has given a new answer now and I like his logic much better than mine.

Edit.

On second inspection I do not think rom's logic for each individual play stands up.

I am still uncertain of my logic though.

Sorry blocked on my browser

XxXTenTacion

When you get the answer can you let us know please. (Answer with logic)

Next attempt.

There are 16 cards. All cards are dealt.

[comment deleted]

16C4*12C4*8C4*1 = 1820*495*70*1 = 63 063 000 ways to choose 4 groups of 4

12C3*9C3*6C3*1    ways to chose 4 groups of 3

12C3*9C3*6C3*1 *4*3*2*1  ways to chose 4 groups of three with no 1s and then add a 1 to each group.

  \(\text{Prob(1 in each group)}=\frac{12C3*9C3*6C3*1 *4!}{16C4*12C4*8C4*1}\\ \text{Prob(1 in each group)}=\frac{12C3*9C3*6C3 *4!}{16C4*12C4*8C4}\\ \text{Prob(1 in each group)}=\frac{369600*4!}{63063000}\\ \text{Prob(1 in each group)}=\frac{3696*4!}{630630}\\ \text{Prob(1 in each group)}=\frac{616*4!}{105105}\\ \text{Prob(1 in each group)}=\frac{14784}{105105}\\ \text{Prob(1 in each group)}=\frac{4928}{35035}\\ \text{Prob(1 in each group)}\approx 0.14065\)

check

((nCr(12,3)*nCr(9,3)*nCr(6,3)*1*4!)/(nCr(16,4)*nCr(12,4)*nCr(8,4)*1) = 0.1406593406593407((nCr(12,3)*nCr(9,3)*nCr(6,3)*1*4!)/(nCr(16,4)*nCr(12,4)*nCr(8,4)*1 = 0.1406593406593407

So I have an answer. I have little idea about whether it is correct.

HINT:

Notice how you can substitute in y= 4x - 19 into 2x + y = 95

 2x + 4x - 19 = 95.

Solve for x

I'm not sure I haven't taken algebra 2 yet, but I think its like htis

Since 2016 is even,

what is (1+i)^2?

What is (1-i)^2

So I think th answer should be 4i, after plugging that in the calc. But EP may be correct he probably took alg 2 already.

Is there any way you can post the image. When I copy pasted the link my browser marked it as illicit ****.    So I can't see it.

My browser blocked the image.