Quite possibly, but it will depend on the lines!
Chat with our AI personalities
Trapezium :Area(A) = 1/2 h(a + b)where a and b are the length of the parallel sidesh= distance between two parallel lines .Perimeter(P) = a + b + c + dwhere a,b,c and d are the length of the sides.
It is possible.
To prove a ring is commutative, one must show that for any two elements of the ring their product does not depend on the order in which you multiply them. For example, if p and q are any two elements of your ring then p*q must equal q*p in order for the ring to be commutative. Note that not every ring is commutative, in some rings p*q does not equal q*p for arbitrary q and p (for example, the ring of 2x2 matrices).
Possibly the letters with no lines of symmetry.
Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.