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LuckyWell [14K]
3 years ago
5

A certain oil refinery keeps intermediate products in 8 tanks. There are 15 pumps of varying capacity that can be assigned to pu

mp the intermediate products from the 8 tanks into a final-product tank. Question: How many ways can you assign the 15 pumps to the 8 tanks so that each tank gets at least one pump?
Mathematics
1 answer:
Furkat [3]3 years ago
4 0

Answer:

Step-by-step explanation:

This question is permutation problem. permutations is the number of arrangements or orderings within a constant group.

The formula for permutation is given as

nPk = n!÷(n-k)!

Where n is the number of objects, k is the is the number of objects we want to chose from.

Solution to the question

15P8 = 15!÷(15-8)!

= 15!÷7!

= 15×14×13×12×11×10×9×8×7×6×5×4×3×2×1÷7×6×5×4×3×2×1

= 259459200

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Please help me due at 12:50. Please please please, I love you . Don’t ignore this please.
bearhunter [10]

Answer:

9/10

Step-by-step explanation:

4/10 + 2/10 + 3/10 = 9/10

4 + 2 + 3 = 9

8 0
3 years ago
CAN SOMEONE PLEASE HELP ME ASAP PLEASEE ANYBODY LITERALLY ANYONE OUT THERE PLEASE. Determine m∠ABQ and determine m∠BCR.
s2008m [1.1K]

Answer:

ABQ  = 139

BCR is the same as QBC since they are alternate interior angles

BCR = 41

Step-by-step explanation:

CBQ and SCB are same side interior angles so they add to 180

2a-9  + 5a +14 = 180

Combine like terms

7a +5 = 180

Subtract 5 from each side

7a = 175

Divide by 7

7a/7 = 175/7

a = 25

ABQ  is the same as SCB  since they are corresponding angles so

ABQ = SCB = 5a+14 = 5*25+14 = 125+14 = 139

BCR is the same as QBC since they are alternate interior angles

BCR = QBC = 2a-9 = 2*25 -9 = 50-9 = 41

4 0
3 years ago
If H be the H.M. between a and b.<br> prove that 1/H-a + 1/H-b = 1/a+1/b​
____ [38]

if susjwmaoakkaajKIJahabsbzh

3 0
3 years ago
Make x the subject of the formula<br> y = (ax)2
katovenus [111]

Answer:

y = (ax)2

y = 2ax

divide both sides by 2a to make x the subject of the formula.

y/2a = 2ax/2a

x = y/ 2a

hope this was helpful,thanks

7 0
2 years ago
How do I convert these probs? into base 10 4120_7, A3B_12, WXYZ_36
natali 33 [55]

4120_7 = 4•7³ + 1•7² + 2•7¹ + 0•7⁰

4120_7 = 4•343 + 1•49 + 2•7 + 0•1

4120_7 = 1372 + 49 + 14

4120_7 = 1435

In base 12, we use the digits 0-9 as well as A for 10 and B for 11. So

A3B_12 = 10•12² + 3•12¹ + 11•12⁰

A3B_12 = 10•144 + 3•12 + 11•1

A3B_12 = 1440 + 36 + 11

A3B_12 = 1487

In base 36, we assign values between 10 and 35 to the letters A-Z, so that

WXYZ_36 = 32•36³ + 33•36² + 34•36¹ + 35•36⁰

WXYZ_36 = 32•46656 + 33•1296 + 34•36 + 35•1

WXYZ_36 = 1492992 + 42768 + 1224 + 35

WXYZ_36 = 1537019

7 0
2 years ago
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