How many times greater is 3,000 than 300

An electronic store is advertising a discount of 18 percent off the inventory. Constantine buys a DVD player that cost $300. Wha

mel-nik [20]

100 % minus 18 % is 82% or .82

So multiply the price by .82 to get the sale price

300(.82)= 246

7 0

3 years ago

According to records, the amount of precipitation in a certain city on a November day has a mean of inches, with a standard devi

Mekhanik [1.2K]

Answer:

The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is mean amount of inches of rain and $\sigma$ is the standard deviation.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

In this question:

Mean $\mu$ , standard deviation $\sigma$

n days:

This means that $s = \frac{\sigma}{\sqrt{n}}$

Applying the Central Limit Theorem to the z-score formula.

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

What is the probability that the mean daily precipitation will be of X inches or less for a random sample of November days?

The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is mean amount of inches of rain and $\sigma$ is the standard deviation.

5 0

3 years ago

Chelsea shows her work in finding the solution to 4x-5=2+3(x-3). After checking her answer in the original equation, she found t

KonstantinChe [14]

Answer:

The solution is x=-2

The mistake could be in operation however here is the all operation

Step-by-step explanation:

Getting each term and sum each one

$4x-5=2+3(x-3)\\4x-5=2+3x-9\\4x-5=-7+3x\\4x-3x=-7+5\\x=-2$