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nordsb [41]
3 years ago
5

The distance covered by one step is 2.5 feet long. how many steps would the average person take to travel 50 feet

Mathematics
2 answers:
Elenna [48]3 years ago
8 0

we know that

using proportion

\frac{1}{2.5} \frac{step}{ft}  =\frac{x}{50}  {ft}  \\\\  50*1=x*2.5 \\\\ x=\frac{50}{2.5} \\\\   x=20\ steps

therefore

<u>the answer is equal to</u>

20\ steps

Stels [109]3 years ago
4 0
20 steps. 50 divided by 2.5.
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vlada-n [284]

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6 0
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Total
balandron [24]

Answer:

Step-by-step explanation:

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5 0
3 years ago
A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent
navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃    to maximize

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij   product size i produced in center j

According to this, we get the following set of variable

X₁₁    product size huge produced in center 1

X₁₂    product size huge produced in center 2

X₁₃   product size huge produced in center 3

X₂₁   product size average produced in center 1

X₂₂   product size average produced in center 2

X₂₃   product size average produced in center 3

X₃₁  product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.-   First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

2.-   Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

3.- Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275

Water available

1.-  22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

2.-  22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

3.-   22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

Demand constrain

Product huge

X₁₁  +  X₁₂  + X₁₃  ≤  710

Product average

X₂₁  + X₂₂ + X₂₃  ≤  900

Product tiny

X₃₁ + X₃₂ + X₃₃  ≤  350

Fraction SP/CC must be the same

First and second centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   =  (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0
3 years ago
You produce 100 items and 47 of them will be shipped out of state. What percent of the items will be shipped out of state?
larisa [96]
47%.
anything that's out of 100 is going to be that number as a percent.
So if you had 34 things being shipped out of state then 34% of them would be shipped out of state.
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3 years ago
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It is recommended to use a Poisson process to model the number of failures in commercial water
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A research paper recommends using Poisson process to model the number of failures in commercial water pipes The paper also gives estimates the failure rate in units of failures per 100 miles of pipe per day; for four different types of pipe and for many different years_ For example, for cast iron pipe in 2005, the authors' estimate is the failure rate is 0.0864 failures per 100 miles per day: Suppose a town had 2500 miles of cast iron pipe underground in 2005 What is the probability of at least two failures
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