-5+(63/9)<br> Can someone help me please.

nevsk [136]

X squared times x squared x to the 4th power
You basically add the exponents

4 0

4 years ago

Plz help asap !!! The high school jazz band is selling homemade leather bracelets at a local craft fair to raise money for a tri

coldgirl [10]

Answer:

b. y=2x-200

c. there will be no profit because 2 times 100 = 200, meaning they only earned back the money they already spent.

d. (domain or y-values) minimum: -200 maximum: 320

(range or x-values) maximum: 260

3 0

3 years ago

Translate the sentence into an inequality.<br> The sum of 8 and y is greater than 26.

SVEN [57.7K]

The sum of 8 and y is greater than 26.

8 + y > 26

6 0

3 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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