i personally chose 0 an my interval
not really
There are no mixed numbers between 0 and 2 with an interval of 18.
Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].
No, it must be a number in the interval [0, 1].
You construct a 95% confidence interval for a parameter such as mean, variance etc. It is an interval in which you are 95 % certain (there is a 95 % probability) that the true unknown parameter lies. The concept of a 95% Confidence Interval (95% CI) is one that is somewhat elusive. This is primarily due to the fact that many students of statistics are simply required to memorize its definition without fully understanding its implications. Here we will try to cover both the definition as well as what the definition actually implies. The definition that students are required to memorize is: If the procedure for computing a 95% confidence interval is used over and over, 95% of the time the interval will contain the true parameter value. Students are then told that this definition does not mean that an interval has a 95% chance of containing the true parameter value. The reason that this is true, is because a 95% confidence interval will either contain the true parameter value of interest or it will not (thus, the probability of containing the true value is either 1 or 0). However, you have a 95% chance of creating one that does. In other words, this is similar to saying, "you have a 50% of getting a heads in a coin toss, however, once you toss the coin, you either have a head or a tail". Thus, you have a 95% chance of creating a 95% CI for a parameter that contains the true value. However, once you've done it, your CI either covers the parameter or it doesn't.
i personally chose 0 an my interval
If the first derivative of a function is greater than 0 on an interval, then the function is increasing on that interval. If the first derivative of a function is less than 0 on an interval, then the function is decreasing on that interval. If the second derivative of a function is greater than 0 on an interval, then the function is concave up on that interval. If the second derivative of a function is less than 0 on an interval, then the function is concave down on that interval.
64
not really
X = (-infinity, 0) U (0, infinity) The above is read as X equals negative infinity, comma zero, union, zero, comma infinity on an open interval (By the way, this interval is made up of two intervals). A parenthesis by a value indicates it is not included. This means X could equal anything between -infinity and 0 and X can equal anything between 0 and infinity. X can not equal -infinity. X can not equal 0. X can not equal infinity. The interval is open because none of the starting or ending values can be a value of X (It's a parenthesis by all the starting and ending values). There is a parenthesis by 0 because 0 is not a possible value of X (the question says so). There is a parenthesis by -infinity and infinity because they are not real numbers. So whether either of them is included in the answer, they always have a parenthesis by them. If a number was included in an interval, there would be a square bracket by it, like this: [ or ]. If the starting number and the ending number on the interval is included then the interval is closed.
There are no mixed numbers between 0 and 2 with an interval of 18.
wha is the interval on a line graph, scale from 0-25?..
Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
Let f(x)=abs(x) , absolute value of x defined on the interval [5,5] f(x)= |x| , -5 ≤ x ≤ 5 Then, f(x) is continuous on [-5,5], but not differentiable at x=0 (that is not differentiable on (-5,5)). Therefore, the Mean Value Theorem does not hold.
No, it must be a number in the interval [0, 1].