Want this question answered?
A variable defined on a continuous interval as opposed to one that can take only discrete values.
A 1-dimensional interval (a, b) is continuous if for any k in (0, 1) the point a + k*(b-a) = a*(1-k) + k*b is also in the interval. This is equivalent to the statement that every point on the line joining a and b is in the interval. The above can be extended to more dimensions analogously.
Yes, it is a Continuous variable measured along an equidistant scale.
The linear discrete time interval is used in the interpretation of continuous time and discrete valued: Quantized signal.
Yes it is because it is a measurement of something usually entering into decimal figures and cannot be simply counted.
Yes.
A variable defined on a continuous interval as opposed to one that can take only discrete values.
A 1-dimensional interval (a, b) is continuous if for any k in (0, 1) the point a + k*(b-a) = a*(1-k) + k*b is also in the interval. This is equivalent to the statement that every point on the line joining a and b is in the interval. The above can be extended to more dimensions analogously.
FXY is a n investment, like buying Japanese Yen, but traded at NYSE in US Dollars. It will go up and down in value like owning Japanese Yen would.
If the function is continuous in the interval [a,b] where f(a)*f(b) < 0 (f(x) changes sign ) , then there must be a point c in the interval a<c<b such that f(c) = 0 . In other words , continuous function f in the interval [a,b] receives all all values between f(a) and f(b)
Yes, it is a Continuous variable measured along an equidistant scale.
why doesn't wiki allow punctuation??? Now prove that if the definite integral of f(x) dx is continuous on the interval [a,b] then it is integrable over [a,b]. Another answer: I suspect that the question should be: Prove that if f(x) is continuous on the interval [a,b] then the definite integral of f(x) dx over the interval [a,b] exists. The proof can be found in reasonable calculus texts. On the way you need to know that a function f(x) that is continuous on a closed interval [a,b] is uniformlycontinuous on that interval. Then you take partitions P of the interval [a,b] and look at the upper sum U[P] and lower sum L[P] of f with respect to the partition. Because the function is uniformly continuous on [a,b], you can find partitions P such that U[P] and L[P] are arbitrarily close together, and that in turn tells you that the (Riemann) integral of f over [a,b] exists. This is a somewhat advanced topic.
The airport code for Forest City Municipal Airport is FXY.
sorry but are gone mad
Interval training is periods of work followed by periods of rest. This is known as work:rest ratio. This is commonly used to train the anaerobic energy system. Continuous training, of which there are many forms does not involve rest periods, although it could involve periods of different intensities (such as Fartlek training).
56
well there is weight training, fartlek,continuous, interval , circuit, flexibility and weight