Write an explicit formula for An, the nth term of the sequence 4, -20, 100, ...

The value of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B0%7D%7B%28m%2Bn%29%7D" id="TexFormula1" title="\frac{0}{(m+n)}" alt="\f

Sever21 [200]

Answer:

two in the middle

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A jar contained 8 balls: 6 red, 1 green and 1 blue.

tester [92]

Answer:

$\{R,GR,BR,GBR,BGR\}$

Step-by-step explanation:

Sample space is a set of possible outcomes of a random experiment.

A random experiment is an experiment whose outcomes can not be predicted.

Number of red balls = 6

Number of green ball = 1

Number of blue ball = 1

Let R denotes red ball, B denotes blue ball and G denotes green ball.

Sample space = $\{R,GR,BR,GBR,BGR\}$

6 0

3 years ago

Jean works for the government and was conducting a survey to determine the income levels of a number of different neighborhoods

NikAS [45]

Answer:

0.0418 = 4.18% probability that the average income level in the neighborhoods was less than $38,000.

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 $s = \frac{\sigma}{\sqrt{n}}$ .