LOT OF POINTS ANSWER ASAP PLS

2^4⋅2^5−(2^2)^2<br><br> PLEASE HELP FAST!

Dvinal [7]

2^9 - 2^4 = 496

when you multiply exponents, you just add the two (4 and 5) and keep the coefficient the same.

when you multiply by an exponent (^2), you multiple the exponents and keep the coefficient the same.

8 0

3 years ago

Please help I will give brainliest only to those who answer respectfully and the best answer

katrin2010 [14]

Answer:

possibly you could use kid per candy, number of trickery treaters or something

you could explain lets say over the holidays Miss. So and So's class's kids got an average of around 4-6 Christmas presents while Miss. Mary Sue's class had kids with 2 presents and kids with 12, since the range between 2-12 is bigger it had a greater standard of deviation

(hopefully this helps and I am not overcomplicating or incorrecting the standard of deviation, good luck)

5 0

3 years ago

In a given year, the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500. If a samp

nika2105 [10]

Answer:

15.15% probability that the sample mean will be $192,000 or more.

Step-by-step explanation:

To solve this problem, 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 $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$