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A(−1, 1), B(2, 3), C(2, 1) 90∘ counterclockwise about vertex A

I don't know if I will get this right, but I'll give it a shot:

So we know that 100 employees have an average of 1.5 cups of coffee a day, and we know the cost is .40 cents per cup, for 5 days a week.

First, we can find out how much it costs for one employee by multiplying 1.5 * 0.40 = 0.60. Which means the cost of one employee per day is 0.60 cents.

And we know that there are 100 employees, so 0.60 * 100 = 60 dollars for all 100 employees.

This means that the company has to pay 60 dollars per day if  all 100 employees wanted to have 1.5 cups of coffee on average per day.

Now, since the company is open 5 days a week all year, we can multiply the amount of weeks by 5: 52 * 5 = 260 days.

Which means you multiply the amount of dollars per day by the amount of days: 260 * 60 = 15600 dollars.

Please respond if I'm incorrect or correct.

Mean = 500        S.D. = 100

Look at positive z-score chart   for   .8500   (this would correspond to 85%  )

     my z-score chart shows   z=1.04  corresponds to  .8508

         so 1.04   standard deviations above the mean  is 85 %

                   1.04 x 100 = 104

                           Mean + 104  =  604    would be the score to be in the upper 15%

For 10 %    find   z-score    .9000    and do similar calc   (closest on my chart is .8997 )

For 2%       find z-score    .9800     and do similar calc   (closest on my chart is .9798 )

Need more information to do anything with this....like the length of the string  AND what is your QUESTION ?   

A rate     1 job / (8hr/d x12 d)   =    1j / 96 hr

Brate      1 job/ (10hr/d x 8d)    =    1j /80 hr

1 job  /  ( 1 j / 96 hr   + 1 j / 80 hr) = 43 .63 hrs  

Together , they complete  8 hr + 10 hr   = 18 hr /day

43.63 hr / 18 hr/day =  2. 42 days

The probability for the first piece is 4/5, and the probability for the second piece is 4/5, so the probability is 16/25.

For x² + 1:   Let x = 2n + 1, where n is any positive integer.   Then  x² + 1 = (2n + 1)² + 1 = 4n² + 2n + 1 + 1 = 2(2n² + n + 1)  which is even, and hence composite (since x starts at 3 when n = 1).  There are therefore an infinite number of positive integers (all the odd numbers > 1) that satisfy the condition.

The two wires form a 75 degree angle at the pole

Law of Cosines    c² = a²  + b²  - 2 a b cos 75       c is the distance we are looking for in the Q    a = 12   b = 16

                                 = 12^2 +16^2 - 2 (12)(16) cos 75

                              c^2 = 300.61

                               c = 17.33 Meters   betwen the wire anchors at the ground

The trapezoid is divided into 5 parts horizontally with AB and CD are equally divided.
The Trapezoid also divided into 5 parts vertically with AD and BC are devided equally.
If the shaded regions K has area 35 and the shaded region L has area 55.
Determine the area of the trapezoid ABCD.

\(\begin{array}{|rcll|} \hline 5b+2*\frac{1}{5}x = 5a &\text{ or }& \boxed{b = a-\frac{2}{25}x} \\ 5c+2*\frac{2}{5}x = 5a &\text{ or }& \boxed{c = a-\frac{4}{25}x} \\ 5d+2*\frac{3}{5}x = 5a &\text{ or }& \boxed{d = a-\frac{6}{25}x} \\ 5e+2*\frac{4}{5}x = 5a &\text{ or }& \boxed{e = a-\frac{8}{25}x} \\ 5f+2x = 5a &\text{ or }& \boxed{f = a-\frac{2}{5}x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{trapezoid } ABCD &=& \left(\dfrac{5a+5f}{2}\right)5h \quad | \quad \boxed{f = a-\frac{2}{5}x} \\ &=& \left(\dfrac{5a+5(a-\frac{2}{5}x)}{2}\right)5h \\ &=& \left(\dfrac{10a-2x}{2}\right)5h \\ &=& (5a-x)5h \\ \mathbf{\text{trapezoid } ABCD} &=& \mathbf{25ah-5xh} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{K:} \\ \hline 35 &=& \left(\frac{d+e}{2}\right)h + 2 \left( \frac{e+f}{2}\right)h \\ 35 &=& \left(\frac{d+e}{2}\right)h + ( e+f )h \\ 35 &=& \left(\frac{a-\frac{6}{25}x+a-\frac{8}{25}x}{2}\right)h + ( a-\frac{8}{25}x+a-\frac{2}{5}x )h \\ \ldots \\ \mathbf{35} &=& \mathbf{3ah-xh} \quad (1) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{L:} \\ \hline 55 &=& 2 \left( \frac{a+b}{2}\right)h + \left(\frac{b+c}{2}\right)h \\ 55 &=& ( a+b )h + \left(\frac{b+c}{2}\right)h \\ 55 &=& ( a+a-\frac{2}{25}x )h + \left( \frac{ a-\frac{2}{25}x + a-\frac{4}{25}x }{2} \right)h \\ \ldots \\ \mathbf{55} &=& \mathbf{3ah-\frac{5}{25}xh} \quad (2) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (2)-(1): & 3ah-\frac{5}{25}xh -(3ah-xh) &=& 55-35 \\ & 3ah-\frac{5}{25}xh -3ah+ xh &=& 20 \\ & -\frac{5}{25}xh + xh &=& 20 \\ & \frac{20}{25}xh &=& 20 \\ & \frac{xh}{25} &=& 1 \\ & \mathbf{xh} &=& \mathbf{25} \\\\ (1): & 3ah &=& 35+xh \\ & 3ah &=& 35+25 \\ & 3ah &=& 60 \\ & \mathbf{ah} &=& \mathbf{20} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{trapezoid } ABCD} &=& \mathbf{25ah-5xh} \\ &=& 25*20-5*25 \\ &=& 500-125 \\ \mathbf{\text{trapezoid } ABCD} &=& \mathbf{375} \\ \hline \end{array} \)

I agree with guest #4:

I do not understand what you are trying to ask.

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Thank you, EpicWater !

cos  ( arcsin (-0.845))

Easy way : 

arcsin (-0.845) = - 57.67°

cos (-57.67°)  ≈ .535 

"Expert" way

We are looking for  the cosine  of an angle whose sine =  -845/ 1000 =  y  /r

Remember that cosine  =  x /r

We need  to  find x  

x =  ±√ [ r^2 - y^2]  = ±√ [ 1000^2 - 845^2 ] = ±534.76 = ±535

This angle is in the 4th quadrant  (the sine is negative there).....but  the cosine is positive in this quadrant

So

cosine  for this angle =  x / r  =  535 / 1000   =  .535

The wires form the hypotenuses of two right triangles.....we need the distance in each case of the opposite side to the 75° angles

sin (75)  =   distance₁ / 12

sin (75)  =   distance ₂ / 16

So....We can find the distance thusly :

(12 + 16) sin (75)  =   distance₁ + distance₂

28sin(75) ≈ 27.05 ft

they were correct, thank you!

im very confused on how to solve for this one, could you check it out as well?

First one

Is not one-to-one

(-pi/2, pi/2)

Second one....Correct  !!!

Thank you so much for your response! Guest. I have contacted my friend and got a much more complex solution, your way may work as well, though! :D

(b) What is the probability that we draw any straight (including "straight flush" and "royal straight flush" hands)?

This is similar  to the last one

If  we let  the ace be the lowest or highest card in the straight.....we have 10 possible straights within the ranks :

A-2-3-4-5

2-3-4-5-6

......

10-J-Q-K-A

So the probability  is just  

C(10,1) * C(4,1)/ C(52,5)  ≈  : .00394  = .394%

(a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1 five (this is one way to get a "straight")?

In each case  we want to  select  1 of 4 cards from each of the specified ranks  = [ C(4,1)]^5  = 4^5

And the total  hands = C(52,5)

So.....the probability is just

4^5 / C(52,5)   ≈  .000394  = .0394%

Note that these hands include the possibility of a straight flush ....