Need help understanding what to do

Mathematics

1 answer:

Juliette [100K]3 years ago

6 0

I believe you are expected to simplify. To do this all you do is group like terms (x, x², x³ etc.) and then simplify the coefficients.

13. 10z + 7z - 19z² - 5z² - 17z

first you can rearrange and group like terms (remember the bring the sign in front of each term with it). I would do it like this:
-19z² - 5z² +10z + 7z - 17z

Now simplify the coefficients:

-24z² + 0z

you can omit the 0z and your simplified answer is

-24z²

Main thing to remember here is that z and z² cannot be simplified together, nor can any variable that has an exponent because the exponent makes them differ.

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Suppose monthly rental prices for a one-bedroom apartment in a large city has a distribution that is skewed to the right with a

omeli [17]

Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

$\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287$