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q6h - every six hours

prn- as needed

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Q: 1 R q6h prn N and V?
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What is pxrxn in math?

Oh, hello there! "Pxrxn" in math usually refers to the product of two numbers. It's like mixing colors on your palette to create a beautiful painting - when you multiply two numbers together, you're creating a new number that is the result of their combination. Just like how each brushstroke adds depth to a painting, each multiplication helps us understand how numbers work together. Just remember to take your time and enjoy the process of learning math - there are no mistakes, just happy little accidents!


Prove that nCr plus nCr minus 1 equals n plus 1Cr?

nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr


Why is it said that poisson distribution is a limiting case of binomial distribution?

This browser is totally bloody useless for mathematical display but...The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]Let n -> infinity while np = L, a constant, so that p = L/nthenP(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)Now lim n -> infinity of (1 - L/n)^n = e^(-L)and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.


What is the general term of the expansion 1-x to the power n?

The r+1 th term is nCr(-x)r where r = 0, 1, 2, ... , n. and where nCr = n!/[r!*(n-r)!]


How do you expand binomials A plus b to the degree of n?

It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n where nCr = n!/[r!*(n-r)!]


1 plus r raised to power n equals?

Binomial Theorem: 1n + nC1*1n-1*r + nC2*1n-2*r2+......+nCn-1*1*rn-1 + rn Or (1+r)n = 1 + n*r + n(n-1)/2! * r2 + n(n-1)(n-2)/3! * r3 + .......... n(n-1)...(n-k)/k! * rk if n < 1 as you cannot calculate the combinations that easily. This gives an accurate approximation provided that abs(x) < 1.


What is the formula for euqated monthly installment of loan amount?

Emi = l * r * ((1 + r)^n / (1 + r)^n - 1) * 1/12 where l = loan amt r = rate of interest n = no of terms


What does the formula n(1-R)L mean?

n(1-R)L is an expression: it is not a formula.


What is the binomial probability formula?

T r+1 = (n / r) (a ^n-r) x (b)^r


How do you find the sum of an infinite geometric series?

An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)


How do you solve for probabilities in binomial distributions?

Suppose you have n trials of an experiment in which the probability of "success" in each trial is p. Then the probability of r successes is: nCr*pr*(1-p)n-r for r = 0, 1, ... n. nCr = n!/[r!*(n-r)!]


What is the formula to find the sum of a geometric sequence?

The formula to find the sum of a geometric sequence is adding a + ar + ar2 + ar3 + ar4. The sum, to n terms, is given byS(n) = a*(1 - r^n)/(1 - r) or, equivalently, a*(r^n - 1)/(r - 1)