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Let's make Z the price of a zucchini, and Z + 0.25 the price of a kiwi.

We know that 3(Z + 0.25) + 4 * Z = 4.95.

If we simplify the equation: 3Z + 0.75 + 4Z = 4.95.

Subtracting 0.75 from both sides: 7Z = 4.20

Z = 0.6 which is the price of a zucchini

Z + 0.25 = 0.85 which is the price of a kiwi. 

Which is what CPhill got!

How are you getting so many of these questions?

So you have 17 wins and 3 losses which means they played 20 games so far.

The question states that it's only 2/3 of all it's games, which means 20 = 2/3. So that means that they played 30 games in total.

3/4 of 30 is 22.5, but we can round up to 23. So that means the team must win 23 games, and lose at MOST 7 games. It already lost 3 games so 7 - 3 = 4.

The remaining games that the team can still lose is 4 games.

Well, I don't really know this but I'll give it a shot:

We need to caculate the amount of ways from (0,0) to (5,7): 12C5=792 ways from (0,0) to (5,7), and then subtract that by the amount of ways that pass the spider.

Next, we also need to caculate the paths from (0,0) to (4,3) and multiply that by the amount of paths from (4,3) to (5,7):

7C3 * 5C1 = 175.

792 - 175 = 617.

We have 4 ways to choose the geography book to occupy the middle position

And  we have  C(5,2)  =  10  ways to choose the  latin textbooks.....and each of these can be arranged in 2 ways

And the other textbooks can be arranged in 6!  = 720 ways

So....the total arrangements  =

4 * C(5,2) * 2 * 720  =

4 * 10 * 2 * 720  =

57,600 possible arrangements

6 minutes 40 sec  =  400 sec

So.....each second the  cold  water fills 1/400 of the tub

8 minutes  =  480 sec

So....each second the  hot water fills  1/480  of the  tub

And  13 min 20 sec  =  800 sec

So....each second, 1/800  of the  tub  is drained

So....we  need to  solve this.....where x is the  number of seconds it takes to  fill the tub

(1/400)x + ( 1/480)x  - (1/800)x   =  1

(2/800)x  + (1/480)x - (1/800)x   = 1

(1/800)x  + (1/480) x  = 1

[ 480 + 800] x / { 800 * 480)   = 1

x  =  (800 * 480)  / { 480 + 800)    =  300 sec  to  fill the  tub =  5 minutes

Binomial Probability:

A:  ₅₀C₂₅·(.44)²⁵·(.56)²⁵    =  0.07813

B:  ₅₀C₂₅·(.44)²⁵·(.56)²⁵ + ₅₀C₂₆·(.44)²⁶·(.56)²⁴ + ₅₀C₂₇·(.44)²⁷·(.56)²³ + ... + ₅₀C₅₀·(.44)⁵⁰·(.56)⁰  =  0.237

Either find a program to do part B or use the normal approximation for the binomial ...    

Call the number of additional weeks  after the  first  five   =  W

And 15 total games have been played during the first  five weeks  = 13 + 2  =15 

The number of total  games  past week 5  =  3W

And the number of total wins past week 5  = 2W

So.....we want to solve this

2  +  2W

________   =   .60              multiply both sides by   13 + 3W

15  + 3W

2 + 2W  =  .60 ( 15 + 3W)      simplify

2 + 2W  =  9  + 1.8W       subtract 1.8W, 2  from both sides

.2W  = 7      divide both sides by .2

W  = 35

So.... 60%  of the games will  be won after week  (5 + 35)   = 40

Check.....after 40 weeks     2 + 2(35)  =  72 games will have been won

And   15 + 3(35)  =  120  games would have been played

So   72 /120    =   .60    = 60%

200 sin 36   +  200 sin 22 =.....

210°  =    30°  reference angle in the  3rd quadrant.......the coordinates  are  ( -√3/2  ,-1/2)

-7pi /4  =   2pi - 7pi/4 =  8pi /4  - 7pi/4 =  pi / 4 = 45°  ....the coordinates are  (√2/2, √2/2)

34pi/3  = 11pi + pi/3  =  240°   = 60°refernce angle in the 3rd quadrant.....the coordinates are  ( -1/2, -√3/2)

Rate * Time  =  Distance

So....you just need to evaluate this :

90 * 3    +  1.5 ( 72)    =      Total Distance

AHHH! i need help too! 

Let the price  of a zucchini  = Z

Then....let the price of a kiwi  = Z  +  .25

And we know that

3 (Z + .25)  +   4(Z)  =  4.95     simplify

4Z +  3Z  +  .75  = 4.95         subtract  .75 from both sides

7Z  =  4.20     divide both sides  by  7

Z =  .60   = the price of a zucchini

Z + .25 =  .85  = the price of a kiiwi

The total possible arrangements  are  9!  = 362800   arrangements

If we  count the ones  where NO  women stand next to each other  ....then note that the women can occupy

positions   1 - 3 - 5 -7   or  2 - 4 -6 - 8   or   3 - 5 -7 - 9  =  3  ways

And for each of these the women can be arranged in 4! = 24 ways

And  for each of these the men can be arranged  in 5!   =120 ways

So....the    number of  arrangements where at least two of the women stand to each other  =

9!  -  (3) (24)(120)  = 

362800 - 8640  =

354,240  possible arrangements

Let the width   = W

Let the length =  W + 3

The perimeter  =  2 ( W + L)

So

58 =  2 ( W  + (W + 3) )      divide both sides by  2

29  = 2W + 3         subtract  3  from both sides

26  = 2W     divide both sides  by  2

13 = W  = the width

And the length  = W + 3   = 16

The volume of a sphere is 523.33 ft cubed. To the nearest foot​, what is the radius of the​ sphere? Use 3.14 for pi.     

The formula is                              V = (4/3) π r³                     

Plug in known values                    523.33  =  (4/3) • 3.14 • r³     

Do a little arithmetic                      523.33  =  4.187 • r³              

Divide both sides by 4.187            128.047  =  r³              

We know that 5 cubed is 125, so the cube root of 128.047 will be slightly larger than 5.   

To get the exact cube root,         

I'll have to use a calculator ..... and the answer is ..... r = 5.04  see below 

When I put 5.04 back into the formula, the answer came out high.  This would be the effect of rounding that I did along the way.  When I work it on the calculator with no rounding, then r³ = 124.999.... that's close enough to 125 to say the radius is 5, therefore r = 5.  

_.

2)

Find a such that ax^2+15x+4 is the square of a binomial.

For this to be true.....we need to have that the discrimant  must  = 0

So

15^2  - 4a (4)   = 0

225  - 16a  = 0

225 =  16a

a =  225 /16

If we take the positive square root of this we have   15/4

And note that   (15/4 x + 2)^2 =  (225/16)x^2 + 2 (15/4)(2)x  + 4  =  (225/16)x^2 + 15 x + 4

See the "answers" here:  https://web2.0calc.com/questions/help_85792#r4

Good luck in choosing the "correct answer" !!.

(1)

(2x+7)(x-5) = -43 + jx       simplify the left side

2x^2 + 7x -10x  - 35  =  -43 + jx          rearrange as

2x^2  + (7 - j)x  + 8  = 0

If this has one real solution.....then the discriminant must   = 0

So

(7 -j)^2  - 4 (2)(8)   =0     

49 - 14j + j^2  - 64  =  0      rearrange as

j^2 -14j - 15  = 0      factor as

(j -15) ( j + 1)   = 0

Set both facotrs to 0   and solve for j  and we get that

j  =15      and j = -1

So....we will have one real solution when  j = -1  or j  =15

See the graph here to confirm this  :  https://www.desmos.com/calculator/dw6lnjpnm7

I'm assuming that you want help with problem #2.

Call the bottom-left vertex A, the bottom-right vertex B, and the top vertex C.

Label the points where the square hits the triangle P, Q, R, and S in this fashion:

-- along AB, let the points of the base be, in order, A, P, Q, and B

-- along BC, let the point of intersection be R, and

-- along AC, let the point of intersection be S.

Given:  AC = 3, CB = 4, and AB = 5

Triangle(APS) is similar to triangle(ACB), so we can call the length of AP = 3x, PS = 4x, and AD = 5x.

Triangle(BQR) is similar to triangle(BCA), so we can call the length of BQ = 4y, BR = 5y, and QR = 3y.

Triangle(CSR) is similar to triangle(CAB), so we can call the length of SR = 5z, CS = 3z, and CR = 4z.

For side(AC), we have AC = AS + SC  =  5x + 3z                   --->            5x + 3z = 3.        (1)

For side(AB), we have AB = AP + PQ + QB  =  3x + 5z + 4y   --->   3x + 4y + 5z  =  5.      (2)

For side(BC), we have BC = BR + RC  =  5y + 4z                   --->            5y + 4z  =  4.      (3)

Since we have a square:  SP  =  SR  =  QR     --->     4x  =  5z  =  3y

Using  5z  =  3y     --->     z  =  3y/5

Substituting this into equation 3:  5y + 4z  =  4   --->   5y + 4(3y/5)  =  4   --->   25y+ 12y  =  20

   --->      37y  =  20     --->     y  =  20/37.

Therefore, since the side of the square is  3y,  side  =  3(20/37)  =  60/37.

Three very "smart" people get 3 different answers to the same question! What does this tell you about "probability"? It tells you that "Probability is a curse upon the world of Mathematics" !! By the way, geno3141's answer =0.9821673525.

Really!   I am astonished!     

Nope! You're correct! Thanks Melody!

I gave you a new answer under Gino's but it is probably not correct.

Andre has a head start     distance = 1 min x 48 m/min = 48 meters

Ursala is swimming (60-48) = 12 meters /m faster            and will catch Andre    48 m / 12 meters /m =  4 minutes after she starts.

From a standard deck of cards, the four Aces, four 2's, four 3's, and four 4's are drawn, forming a smaller deck of 16 cards. All 16 cards are dealt at random to four players so that each player gets four cards. What is the probability that each player gets an Ace?

I have not looked at Geno's answer, I just decided to think about it independently.

I certainly will not promise mine is correct.

There are   16C4*12C4*8C4*4C4 ways to choose the hands but I this will be double counting so I think I need to divide by

4!

=63063000/4!

=2627625   total selection of hands (maybe)

Take all the Aces out and give three cards to each

(12C3*9C3*6C3) /4!*4!  maybe = 369600

369600/2627625

369600/2627625 = 0.1406593406593407 = 64/455     (probably wrong)

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Tell us when you know the answer, preferably with working:)