I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

$W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}$

The values of the cumulative distribution function of W, denoted by $\phi$ , can be found in the attached file. Lets call $y = \frac{z-70}{4}$ . We have

$0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)$

Thus

$\phi(y) = 1-0.1 = 0.9$

by looking at the table, we find that y = 1.28, therefore

$\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12$

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .

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8 0

3 years ago

What is continuous compounding? how does the apy for continuous compounding compare to the apy for, say, daily compounding? e

aliya0001 [1]

Continuous compounding is the mathematical limit that compound interest can reach.

It is the limit of the function A(1 + 1/n) ^ n as n approaches infinity. IN theory interest is added to the initial amount A every infinitesimally small instant.
The limit of (1 + 1/n)^n is the number e ( = 2.718281828 to 9 dec places).

Say we invest $1000 at daily compounding at yearly interest of 2 %. After 1 year the $1000 will increase to:-

1000 ( 1 + 0.02/365)^365 = $1020.20

with continuous compounding this will be

1000 * e^1 = $2718.28

4 0

3 years ago

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Select the true statement. A. Neither function is even. B. Only function f is even. C. Only function g is even. D. Both function

astraxan [27]

Answer:

The answer is "Option C".

Step-by-step explanation:

The graph of the even function will have both ends pointed in the same direction, while the graph of the odd function will have both ends pointed inside the reverse way. Furthermore, as the y-axis approaches, and also regular expression graph becomes symmetric. The graph of an unusual function, on either hand, is symmetric around the origin and also has rotational symmetry around the origin. They could see that the graph of function g meets a function requirement by looking at it. That graph of function f, but on the other hand, is not symmetric about its y-axis, implying that function f is not even. As both a result, g is a function that is even.

5 0

3 years ago

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