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12

The percentage is given by  68 / 83  ≈  .82

13   Order the data      

Group A      32   36   36  37  41  42  44  45

Group B      26   36   36  38  40  40  40  42     

Mean Group A   =   39.125

Mean Group B   =   37.25       so....not true  

Range

Group A  =  45 - 32  =  13

Group B  =  42  - 26  = 16    false

Mode

Group A  = 36

Group B  = 40   false

Median 

Group A  =  [ 37 + 41] / 2  =  78/ 2  =  39

Group B  = [ 38 + 40 ] / 2  = 78 /    = 39    true

This is what I did to get 85)

Volume of Hemisphere: \((2/3) \Pi r^3=(2/3)\Pi 3^3=56.5486677 \)

Volume of cylinder is \(\Pi r^2h=\Pi 3^2*12=339.29 \)

339.29+56.5486677=395.8 in.^3 (rounded to the nearest tenth) 

25%  pay   $600 - $649

5%  pay   < $600

So....about  30% pay less than  $650 

Thanks you guys!

ACB will be an inscribed angle......it it measures 70°, then the arc it intercepts (AB) has twice this measure = 140°

Then....the length of this minor arc is given by :

Arc Length  =  2 * pi * radius * (140 /360)  =

2 * pi * 18 *  ( 7/18)  =

2 * pi * 7   =

14 pi   units

Good Job, Guest  !!!!

85

The cylinder (and hemisphere) both have a radius of 6....so....the total volume =

Volume of a cylinder  with a radius of 6 and height of 12   +  Volume of a hemisphere with a radius of 6

So  we have

pi  [  r^2 * height of cylinder   +    (1/2)(4/3) r^3  ]  =

pi  [ 6^2 * 12  +  (2/3)*6^3 ]  =

pi [ 432  + 144 ]  =

pi  [ 576 ]  ≈   1809.6 in ^3

I will take a crack at this! CPhill: Please check this. Thanks.

Let us designate the center of the circle as D

If the angle ACB was subteteded to the center of the circle D, then the formed angle ADB = 2 x angle ACB(70) = 140 degrees.

Will convert 140 degrees to radians =140Pi / 180 =7Pi/9, then the length of the arc AB=Angle ADB x Radius =7Pi/9 x 18 = 14Pi

Since the tangent is negatie and the csc is positive....  " t "  must fall into the second quadrant

tan t  =  -5 / 1    =   5 / - 1 =   y / x

We need to find r   =  √  [ x^2 + y^2]  =  √ [ (-1)^2  + 5^2 ]  = √26

sin t  =  y / r  =  5 / √26  =   (5/26)√26

cos t  =  x / r  =  -1 / √26  =  -√26/26

csc t  =   r / y  =  √26 / 5

sec t   =  r / x =  √26 / - 1    =  -√26

cot t  =  x / y  =   -1/5

And there you go  !!!

84

Volume of cylinder  =   pi *8^2 * 16  =  1024 pi  in^3     (1)

Volume of cone  = (1/3) pi * 4^2 * height  =  16/3 pi * height    (2)

If they have equal volumes.....set (1) and (2) equal  and solve for the cone's height

1024 pi    =  (16/3) pi * height        divide out pi

1024  = (16/3) * height       multiply both sides by 3/16

1024 (3/16)  = height 

192  in  = height

So.....the cone is    192/16  =  12 times as high as the cylinder

I need help with all, but any help is great help!!! Thanks in advance! 

Thanks man!

80

The  oblique cylinder doesn't matter.....we can consider it to be just like a "normal" cylinder as far as solving for the radius...so we have...

Volume of cylinder  = pi * r^2 * height

36 pi  = pi * r^2 * 18       divide both sides by 18 pi

2 = r^2      take the square root of both sides

√2  = r  ≈ 1.4 in

79

x^6 - 2x^5 - x^3 - 7  -  (3x^6 - 3x^5 + 2x^2 - 5)

Distribute  the  " - '   across the terms in the parentheses

x^6 - 2x^5 - x^3 - 7 - 3x^6 + 3x^5 - 2x^2 + 5       combine like terms

x^6 -3x^6 - 2x^5 + 3x^5 - x^3 - 2x^2 - 7 + 5

-2x^6 + x^5 - x^3 - 2x^2 - 2

Christ! HSC, Are you off your meds again?!!

78.  Volume of prism  = Area of Base * Height  

The base is a square, so its area is  6^2  = 36 in^2

The height  is 10 in

So the volume is   36 * 10  =  360 in^3

Volume of cylinder = pi * (diameter/ 2)^2 * height  = pi * (10/2)^2 * 6  = pi * 5^2 * 6  =

150pi in^3  ≈ 471.23 in ^3

So.....the cylinder has the greater volume

You can enter both e and pi from the calculator on the home page here (top line near the left hand side).

Yes, Imeant B.

thanks!! 

for every proper integer factor of n that is not n^1/2 there exists another proper integer factor of n such that their product is n

proof: if the proper factor is a, just divide n by a- n/a is a different proper factor, and a*(n/a)=n

this means that we can take the factors of n that are not the square root of n and pair two factors together if their product is n.

Suppose we have exactly K proper factors that are not the square root of n- this means that we have exactly K/2 (meaning that K is divisible by 2) pairs of proper factors. So when we'll multiply those factors of n (except for sqrt(n) of course) we'll get n^K/2.

why? because instead of multiplying all of the factors together, we can first find the product of each of the pairs, and then multiply the results together. but the product of each of the pairs is exactly n, and we have exactly K/2 pairs, therefore the answer is n^K/2.

note- i did not assume that sqrt(n) is a factor of n- when n is not a perfect square, sqrt(n) is not a factor of n at all (in fact, it's irrational). But when sqrt(n) IS a factor of n, we cannot pair him like the other factors, because his "partner" will have to be n/sqrt(n)=sqrt(n), but because the members of a pair have to be different, i cannot allow that

so, when n IS NOT a perfect square, sqrt(n) is not a factor of n, meaning that K=x (x=number of proper positive divisors)

, therefore the product of the proper positive divisors of n is n^x/2. We already know that the product is n^(ax+b)/c, meaning that:

x/2=(ax+b)/c /multiply by c:

c*(x/2)=ax+b/ subtract ax from both sides:

(x/2)(c-2a)=b

Now, there is no single value for a+b+c- for example if x=6, a=-1, b=21 and c=5 we get x/2=(ax+b)/c and a+b+c=25

but a=1 b=0 and c=2 satisfies x/2=(ax+b)/c too, except this time a+b+c=3.

I'm not 100% sure i understood your question, are the numbers a b and c constants that satisfy n^(ax+b)/c=product of proper factors for every n, or just for a specific n? do a and b have to be positive?

i'm going to assume that you didn't mean that a b and c are constants, because it makes no sense.

but even when a, b, and c are have to be non-negative integers your question doesn't make sense.

i think i'll stop here.

please correct your question.

83)

Volume of cube = S^3, where S = one side

V = 16^3 =4,096 mm^3 - volume of the cube

Volume of square pyramid =S^2 x h/3, where S=side, h=height

V =S^2 x h/3

4,096 =16^2 x h/3

4,096 = 256 x h/3 Divide both sides by 256

16 = h/3 Cross multiply

h = 48 mm - Height of the pyramid

48 / 16 = 3x - higher is the square pyramid than the cube.

81)

The answer is "B"

79)

Simplify the following:

-(3 x^6 - 3 x^5 + 2 x^2 - 5) + x^6 - 2 x^5 - x^3 - 7

-(3 x^6 - 3 x^5 + 2 x^2 - 5) = -3 x^6 + 3 x^5 - 2 x^2 + 5:

-3 x^6 + 3 x^5 - 2 x^2 + 5 + x^6 - 2 x^5 - x^3 - 7

Grouping like terms, x^6 - 3 x^6 + 3 x^5 - 2 x^5 - x^3 - 2 x^2 - 7 + 5 = (x^6 - 3 x^6) + (-2 x^5 + 3 x^5) - x^3 - 2 x^2 + (-7 + 5):

(x^6 - 3 x^6) + (-2 x^5 + 3 x^5) - x^3 - 2 x^2 + (-7 + 5)

x^6 - 3 x^6 = -2 x^6:

-2 x^6 + (-2 x^5 + 3 x^5) - x^3 - 2 x^2 + (-7 + 5)

3 x^5 - 2 x^5 = x^5:

-2 x^6 + x^5 - x^3 - 2 x^2 + (-7 + 5)

5 - 7 = -2:

-2 x^6 + x^5 - x^3 - 2 x^2 + -2

Factor -1 out of -2 x^6 + x^5 - x^3 - 2 x^2 - 2:

-(2x^6 - x^5 + x^3 + 2x^2 + 2)

Thanks so much! 

fortnite is lit 

Just use Combinations of 15 students chosen 4 at a time:

15C4 =1,365 ways to choose a team of 4 students.

Nvm I got it! 15*14*13*12=32760

4*3*2*1=24

32760/24=1365

1/1365