Which numbers are irrational or rational?

Mathematics

1 answer:

My name is Ann [436]4 years ago

5 0

Answer:

Rational:

Square root of 81

2 1/3

3.4

-9

- 3/10

Irrational:

3.3333333...

Step-by-step explanation:

Rational numbers terminate or end, irrational don't.

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G(a)= a^6- 5a f(a)=–2a +5 Find g(f(a))

Nina [5.8K]

Answer:

G ( − 2 a + 5 ) = 64 a^ 6 − 960 a ^5 + 6000 a^ 4 − 20000 a^ 3 + 37500 a ^2 − 37490 a + 15600

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A high-profile consulting company chooses its new entry-level employees from a pool of recent college graduates using a five-ste

Paladinen [302]

Answer:

Only 0.0668 of the candidates, or 6.68%, have a GPA above 3.9.

Step-by-step explanation:

The GPA, according to the question, is a random variable normally distributed. A normal distribution is determined by its parameters. These parameters are the population mean, $\\ \mu$ , and the population standard deviation, $\\ \sigma$ . The normal distribution for GPA has a $\\ \mu = 3.3$ , and $\\ \sigma = 0.4$ .

To find probabilities related to normally distributed data, we can use the standard normal distribution. To use it, we first need to find the corresponding standardized score or z-score for the value we want to obtain the probability. This value, in this case, is x = 3.9. The related z-score or standardized value is given by the formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

A z-score tells us the distance from the mean in standard deviations units. A negative value indicates that the value is below the mean. A positive value is, conversely, above the mean.

And we already have all the necessary data to use [1].

Thus

$\\ z = \frac{3.9 - 3.3}{0.4}$

$\\ z = \frac{0.6}{0.4}$

$\\ z = 1.5$

This value for z tells us that the "equivalent" raw score, x = 3.9, is above the mean, and it is at 1.5 standard deviations units from the population mean.

We can find the probabilities for standardized values in the cumulative standard normal table, available on the Internet or in Statistic books (we can also use statistical packages or even spreadsheets).

To use the cumulative standard normal table, we have the z-score as an entry. With this value, we find the z column on this table, and, since it is z = 1.5, we only need to select the first column (which has two decimal places, that is, .00). Then, the cumulative probability for P(z<1.50) = 0.9332.

However, we are asked for the percent of candidates that have a GPA above 3.9 (z-score = 1.50). This probability is the complement of P(z<1.50) or 1 - P(z<1.50). Mathematically

$\\ P(z3.9) = 1$