Answer:
97.06
Step-by-step explanation:
5047 / 52
= 97.06
There are many equations that has a constant of
proportionality, might want to make it more specific.
Answer:
Below are the responses to the given question:
Step-by-step explanation:
Let X become the random marble variable & g have been any function.
Now.
For point a:
When X is discreet, the g(X) expectation is defined as follows
Then there will be a change of position.
E[g(X)] = X x∈X g(x)f(x)
If f is X and X's mass likelihood function support X.
For point b:
When X is continuing the g(X) expectations is calculated as, E[g(X)] = Z ∞ −∞ g(x)f(x) dx, where f is the X transportation distances of probability.If E(X) = −∞ or E(X) = ∞ (i.e., E(|X|) = ∞), they say it has nothing to expect from EX is occasionally written to stress that a specific probability distribution X is expected.Its expectation is given in the form of,E[g(X)] = Z x −∞ g(x) dF(x). , sometimes for the continuous random vary (x). Here F(x) is X's distributed feature. The anticipation operator bears the lineage of comprehensive & integral features. The superposition principle shows in detail how expectation maintains equality and is a skill.
The function will enter the graph graph in the upper left hand region and exit in the upper right hand region and overall the graph will be concave upwards.
For determining the end behavior of a polynomial, there is just 2 things to take notice of.
1. Is the leading coefficient positive or negative?
2. Is the degree of the polynomial odd or even?
For odd ordered polynomials, the curve starts in either quadrant II or III, and ends in quadrant IV, or I. Basically, if it's positive, the curve enters the graph somewhere in the lower left hand region, and exits the graph in the upper right hand region. If the coefficient is negative, it enters in the upper left hand region, and exits in the lower right hand region.
For even ordered polynomials, the graph is either concave upwards (positive leading coefficient) or concave downwards (negative leading coefficient).
In this problem, 14 is an even number and since the coefficient is positive, the function will enter the graph graph in the upper left hand region and exit in the upper right hand region and overall the graph will be concave upwards.
Answer:
p(z<0.42) = 0.6628
Step-by-step explanation:
A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. That is to say:
μ = 0
σ² = 1
Using a calculator, we find that:
p(z<0.42) = 0.6628 (See picture attached)
