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ki77a [65]
3 years ago
8

Maren is painting some doors that are all the same size. She used 3 liters of paint to cover 1 4/5 doors. How many liters of pai

nt are needed for 1 door?
Mathematics
1 answer:
Wewaii [24]3 years ago
5 0

Answer:5/3 litres of paint is needed for 1 door.

Step-by-step explanation:

All the doors that Maren is painting are of the same size.

She used 3 liters of paint to cover 1 4/5 doors. Converting 1 4/5 into improper fraction, it becomes 9/5.

Let the number of litres of paint needed to paint 1 door would be x. Therefore

3 litres = 9/5 doors

x litres = 1 door

Crossmultiplying, it becomes

3 × 1 = x × 9/5

3 = 9x/5

9x = 3 × 5 = 15

x = 15/9 = 5/3 litres

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Let P = 0.50.30.50.7 be the transition matrix for a Markov chain with two states. Let x0 = 0.50.5 be the initial state vector fo
pav-90 [236]

Answer:

Probability distribution vector = \left(\begin{array}{c}0.375\\ 0.625 \end{array} \right)

Step-By-Step Explanation

If P=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right)  is the transition matrix for a Markov chain with two states.  

x_{0}=\left(\begin{array}{c}0.5\\ 0.5 \end{array} \right)  be the initial state vector for the population.

X_{1}=P x_{0}=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right) \left(\begin{array}{c}0.5\\ 0.5 \end{array} \right) =\left(\begin{array}{c}0.4\\ 0.6 \end{array} \right)  

X_{2}=P^{2} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)  

X_{3}=P^{3} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)  

X_{30}=P^{30} x_{0}=\left(\begin{array}{c}0.37499\\ 0.625 \end{array} \right)  

In the long run, the probability distribution vector Xm approaches the probability distribution vector \left(\begin{array}{c}0.375\\ 0.625 \end{array} \right) .

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4 0
3 years ago
HELP ASAP!!! Alex bought a notebook containing 96 pages, and numbered them from 1 through 192. Bob tore out 25 pages of Alex’s n
zubka84 [21]
<h3>Answer: No it is not possible</h3>

=====================================================

Explanation:

Page 1 is labeled with 1 and 2, which sum to 1+2 = 3

Page 2 is labeled with 3 and 4 which sum to 3+4 = 7

Page 3 is labeled with 5 and 6 which sum to 5+6 = 11

and so on until we reach

Page 96 is labeled with 191 and 192, which sum to 191+192 = 383

Note how each page has an odd page number label and an even number label (odd on the front side; even on the back side). Adding any odd number to an even number will result in an odd number. We can prove it as such

x = some odd number = 2m+1, m is any integer

y = some even number = 2n, n is an integer

x+y = 2m+1+2n = 2(m+n)+1 = some other odd integer because it is in the form 2p+1 with p = m+n as an integer

This explains why the results 3,7,11,..,383 are all odd.

------------------------

So we effectively have this set of values {3,7,11,...,383}. This is an arithmetic sequence with 3 as the first term and 4 as the common difference.

If we add two odd numbers together, we get an even number (proof is similar to one shown above)

odd + odd = even

But if we add in another odd number, then we'll go back to an odd result

odd + odd + odd = odd

If we have an odd number of odd numbers added up, then the result will be odd. In this case, we're adding 25 values from the set {3,7,11,...,383}. The value 25 is odd, so we have an odd number of values from  {3,7,11,...,383} being added up. Therefore, the result Bob will get will always be odd. There is no way to get a sum of 2012 because this value is even.

3 0
3 years ago
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