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3 years ago

What is the domain and range for the following function and its inverse f(x) = x^2 – 5

Mathematics

2 answers:

RSB [31]3 years ago

8 0

Answer:

Step-by-step explanation:

hello :

the domain is : D = R = =]-∞;+∞[ ( all numbers reals)

let : f(x) = y

y = x² - 5

x² = y+5

x exist : y + 5 ≥ 0

y ≥ - 5

so : the inverse is [-5;+∞[

Mice21 [21]3 years ago

4 0

for parabolas, the domain will always be real numbers.

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3 years ago

Read 2 more answers

Of the total population of American households, including older Americans and perhaps some not so old, 17.3% receive retirement

Alex Ar [27]

Answer:

47.54% probability that more than 20 households but fewer than 35 households receive a retirement income

Step-by-step explanation:

We use the binomial aproxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.173, n = 120$ . So

$\mu = E(X) = np = 120*0.173 = 20.76$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.173*0.827} = 4.14$

In a random sample of 120 households, what is the probability that more than 20 households but fewer than 35 households receive a retirement income?

We are working with discrete values, so this is the pvalue of Z when X = 35-1 = 34 subtracted by the pvalue of Z when X = 20 + 1 = 21.

X = 34

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{34 - 20.76}{4.14}$