What is the diameter of a circle with an area of 100 pie square feet

A gas has a volume of 5.0 L at a pressure of 50kPa. What happens to the volume when then the pressure is increased 125kPa. The t

Olegator [25]

9514 1404 393

Answer:

volume goes to 2.0 L

Step-by-step explanation:

At constant temperature, volume and pressure are inversely related. Increasing the pressure by a factor of 125/50 = 2.5 will cause the volume to be reduced by that factor.

volume = 5.0L / 2.5

volume = 2.0 L

7 0

2 years ago

Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose

spayn [35]

Answer:

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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 normal distributions 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.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$