A negative times a negative is a positive. A simple way to remember this is... n * p = n p* n = n p * p = p n * n = p There will always be two negatives, and two positives.
If: m = n+x/p then x = p(m-n)
Simple Interest = p * i * n p is principle and i is interest rate per period and n is the number of periods. A = P(1 + r)n is for compound interest.
Positive plus positive is positive. Negative plus negative is negative. Positive plus negative is positive if the absolute value of the positive number is greater than the absolute value of the negative one. Positive plus negative is negative if the absolute value of the negative number is greater than the absolute value of the positive one.
P V = n R TDivide each side by ( n T ):(P V) / (n T) = R
P= positive N=negative P x N = N N x P = N P x P = P N x N = P Hope that helps!?!?!
N - p% = N - p% of N = N*(1 - p%) = N*(1 - p/100) or N*(100 - p)/100
BJT is nothing but the addition of two PN junction diodes. There are two types of BJT= P-N-P or N-P-N P-N N-P + or + N-P = P-N-P P-N =N-P-N SCR is a thyristor which is made adding two BJTs. Of course they are made of sillicon. Exempli gratia: P-N-P + + N-P-N = P-N-P-N comparison between scr bjt and mosfet Check the related link for further information.
The answer is negative four BECAUSE... 20/5 is POSITIVE four 20/-5 is a NEGATIVE four because a positive divided by a negative is a negative. Easy way to remember negs/pos: n * p = n p* n = n p * p = p n * n = p n / p = n p / n = n p / p = p n / n = p There are always two ways to get a positive, and two ways to get a negative. Very simple.
Positive. p*p=p p*n=n n*n=p
A negative times a negative is a positive. A simple way to remember this is... n * p = n p* n = n p * p = p n * n = p There will always be two negatives, and two positives.
What is N E P?
Multiplying by a negative changes the sign of the original number. P x N = N N x N = P Multiplying by a positive keeps the sign of the original number. P x P = P N x P = N
n-p-n and p-n-p junction diodes
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
Mobil-1 p/n M1-103 Wix p/n 51394 Amzoil p/n EA15K09 K & N p/n KN-128 Fram p/n PH4967
We have to use the expression P(X=x) = nCx px (1--p)(n--x) Here n = n and p=p and x = 1 or x>1 P(X>/=1) = 1 -- P(X</=1) So, P(X<=1) = P(X=0) + P(X=1) This gives nC0 p0 (1--p)(n-0)+ nC1 p1 (1--p)(n--1) ie (1--p)n + n p (1--p)(n--1)