X is at most 30 represent as an algebraic inequality

How many unique 5 card hands can be drawn from a deck of cards?

sweet-ann [11.9K]

Answer:

0 2.598.960

Step-by-step explanation:

There are 52 different cards. Order does not matter.

$\dfrac{52!}{(52 - 5)!5!} = 2598960$

7 0

3 years ago

A batch of 40 semiconductor chips is inspected by choosing a sample of 5 chips. Assume 10 of the chips do not conform to custome

butalik [34]

Answer:

a) 658008 samples

b) 274050 samples

c) 515502 samples

Step-by-step explanation:

a) How many ways sample of 5 each can be selected from 40 is just a combination problem since the order of selection isn't important.

So, the number of samples = ⁴⁰C₅ = 658008 samples

b) How many samples of 5 contain exactly one nonconforming chip?

There are 10 nonconforming chips in the batch, and 1 nonconforming chip for the sample of 5 be picked from ten in the following number of ways

¹⁰C₁ = 10 ways

then the remaining 4 conforming chips in a sample of 5 can be picked from the remaining 30 total conforming chips in the following number of ways

³⁰C₄ = 27405 ways

So, total number of samples containing exactly 1 nonconforming chip in a sample of 5 = 10 × 27405 = 274050 samples

c) How many samples of 5 contain at least one nonconforming chip?

The number of samples of 5 that contain at least one nonconforming chip = (Total number of samples) - (Number of samples with no nonconforming chip in them)

Number of samples with no nonconforming chip in them = ³⁰C₅ = 142506 samples

Total number of samples = 658008

The number of samples of 5 that contain at least one nonconforming chip = 658008 - 142506 = 515502 samples

6 0

3 years ago

Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.

love history [14]

Answer:

x3+y3+z3=k with k is integer from 1 to 100

solution x=0 , y=0 and z=1 and k= 1

For K= 1 , we have the following solutions (x,y,x) = (1,0,0) ; or (0,1,0) ; or (0,0,1) ,

For k =1 also (9,-8,-6) or (9,-6,-8) or (-8,-6,9) or (-8,9,-6) or (-6,-8,9) or (-6,9,8)

And (-1,1,1) or (1,-1,1)

=>(x+y)3−3x2−3xy2+z3=k

=>(x+y+z)3−3(x+y)2.z−3(x+y).z2=k

=>(x+y+z)3−3(x+y)z[(x+y)−3z]=k

lety=αand z=β

=>x3=−α3−β3+k

For k= 2 we have (x,y,z) = (1,1,0) or (1,0,1) or (0,1,1)

Also for (x,y,z) = (7,-6,-5) or (7,-6,-5) or (-6,-5,7) or (-6,7,-5) or (-5,-6,7) or (-5,7,-6)

For k= 3 we have 1 solution : (x,y,z) = (1,1,1)

For k= 10 , we have the solutions (x,y,z) = (1,1,2) or (1,2,1) or (2,1,1)

For k= 9 we have the solutions (x,y,z) = (1,0,2) or (1,2,0) or (0,1,2) or (0,2,1) or (2,0,1) or (2,1,0)

For k= 8 we have (x,y,z) = ( 0,0,2) or (2,0,0) or (0,2,0)

For k= 17 => (x,y,z) = (1,2,2) or (2,1,2) or ( 2,2,1)

For k = 24 we have (x,y,z) = (2,2,2)

For k= 27 => (x,y,z) = (0,0,3) or (3,0,0) or (0,3,0)

for k= 28 => (x,y,z) = (1,0,3) or (1,3,0) or (1,3,0) or (1,0,3) or (3,0,1) or (3,1,0)

For k=29 => (x,y,z) = (1,1,3) or (1,3,1) or (3,1,1)

For k = 35 we have (x,y,z) = (0,2,3) or (0,3,2) or (3,0,2) or (3,2,0) or 2,0,3) or (2,3,0)

For k =36

we have also solution : x=1,y=2andz=3=>

13+23+33=1+8+27=36 with k= 36 , we have the following

we Have : (x, y,z) = (1, 2, 3) ; (3,2,1); (1,3,2) ; (2,1,3) ; (2,3,1), and (3,1,2)

For k= 43 we have (x,y,z) = (2,2,3) or (2,3,2) or (3,2,2)

For k = 44 we have ( 8,-7,-5) or (8,-5,-7) or (-5,-7,8) or ( -5,8,-7) or (-7,-5,8) or (-7,8,-5)

For k =54 => (x,y,z) = (13,-11,-7) ,

for k = 55 => (x,y,z) = (1,3,3) or (3,1,3) or (3,1,1)

and (x,y,z) = (10,-9,-6) or (10,-6,-9) or ( -6,10,-9) or (-6,-9,10) or (-9,10,-6) or (-9,-6,10)

For k = 62 => (x,y,z) = (3,3,2) or (2,3,3) or (3,2,3)

For k =64 => (x,y,z) = (0,0,4) or (0,4,0) or (4,0,0)

For k= 65 => (x,y,z) = (1,0,4) or (1,4,0) or (0,1,4) or (0,4,1) or (4,1,0) or (4,0,1)

For k= 66 => (x,y,z) = (1,1,4) or (1,4,1) or (4,1,1)

For k = 73 => (x,y,z) = (1,2,4) or (1,4,2) or (2,1,4) or (2,4,1) or (4,1,2) or (4,2,1)

For k= 80=> (x,y,z)= (2,2,4) or (2,4,2) or (4,2,2)

For k = 81 => (x,y,z) = (3,3,3)

For k = 90 => (x,y,z) = (11,-9,-6) or (11,-6,-9) or (-9,11,-6) or (-9,-6,11) or (-6,-9,11) or (-6,11,-9)

k = 99 => (x,y,z) = (4,3,2) or (4,2,3) or (2,3,4) or (2,4,3) or ( 3,2,4 ) or (3,4,2)

(x,y,z) = (5,-3,1) or (5,1,-3) or (-3,5,1) or (-3,1,5) or (1,-3,5) or (1,5,-3)

=> 5^3 + (-3)^3 +1 = 125 -27 +1 = 99 => for k = 99

For K = 92

6^3 + (-5)^3 +1 = 216 -125 +1 = 92

8^3 +(-7)^3

Step-by-step explanation:

4 0

3 years ago

Nik needs to estimate how many books will fit in a bin. Each book is 0.75 feet tall, 0.75 feet wide, and 0.25 feet thick. The bi

cricket20 [7]

Step-by-step explanation:

first you find the volume of a bin.

that is multiplication of 3 dimensions given.

4*4*4=64 cubic feet...

now volume of books..

0.75*0.75.0.25=0.14 cubic feet

now the number of books that will fit the bin=volume of bin/volume of books

64/0.14=455

455 books

7 0

2 years ago

I need help..

Debora [2.8K]

This is exactly like the one with the 'n's that you posted yesterday.
I guess you didn't get enough help on that one to understand it.

7k/8 - 3/4 - 5k/16 = 3/8

3/4 is the same as 6/8.
Add it to each side of the equation:

7k/8 - 5k/16 = 9/8

Multiply each side by 16 :

14k - 5k = 18

Add up the 'k's on the left side:

9k = 18

From this point, you can proceed directly to the numerical value of 'k'
if you need it.

In your question, you said you need help, and I showed you how to
strip the problem down so that the only thing left is ' k = something '.
Giving you the answer is no help. Nobody actually needs the answer.

8 0

3 years ago

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