A function is positive on an interval, say, the interval from x=a to x=b, if algebraically, all the y-coordinate values are positive on this interval; and graphically, the entire curve or line lies above the x-axis.on this interval.
-15 < x
(-3,3)
the ansewer to it is to do y frist the x why b<3
n order to fit the Poisson distribution, we must estimate a value for λ from the observed data. Since the average count in a 10-second interval was 8.392, we take this as an estimate of λ (recall that the E(X) = λ) and denote it by ˆλ.
up and down. the x goes left and right
A confidence interval of x% is an interval such that there is an x% probability that the true population mean lies within the interval.
Interval (hrs) = M2 phase (degrees) x M2 period (hrs) / 360 (degrees) Interval = 225 x 0.78 / 360 Interval = 0.4875hrs
A function is positive on an interval, say, the interval from x=a to x=b, if algebraically, all the y-coordinate values are positive on this interval; and graphically, the entire curve or line lies above the x-axis.on this interval.
One.
Generally speaking an x% confidence interval has a margin of error of (100-x)%.
Let g(x) = interval [0, x] of sin t dt, and f(t) = sin t. Since f(t) is a continuous function, the part one of the Fundamental Theorem of Calculus gives, g'(x) = sin x = f(x) (the original function). If you are interested in the interval [x, 0] of sin t dt, then just put a minus sign in front of the integral and interchange places of 0 and x. So that, g(x) = interval [x, 0] of sin t dt = -{ interval [0, x] of sin t dt}, then g'(x) = - sin x.
-15 < x
what exponential function is the average rate of change for the interval from x = 7 to x = 8.
The confidence interval consists of a central value and a margin of error around that value. If it is an X% confidence interval then there is a X% probability that the true value of the statistic in question lies inside the interval. Another way of looking at it is that if you took repeated samples and calculated the test statistic each time, you should expect X% of the test statistics to fall within the confidence interval.
(-3,3)
f(x) is decreasing on the interval on which f'(x) is negative. So we want: (x2-2)/x<0 For this to be true either the numerator or the denominator (but not both) must be negative. On the interval x>0, the numerator is negative for 0<x<sqrt(2) and the denominator is positive for all x>0. On the interval x<0, the denominator is negative for all values on this interval. The numerator is positive on this interval for x<-sqrt(2). So, f' is negative (and f is decreasing) on the intervals: (-infinity, -sqrt(2)), (0, sqrt(2))