If you know any of these could yoy hep me out :D

liberstina [14]

Answer: a) -240

<u>Step-by-step explanation:</u>

$2\sqrt{-32}\cdot 5\sqrt{-18}\\\\=2\cdot 5\sqrt{-32\cdot -18}\\\\=10\sqrt{-1\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot -1\cdot 2\cdot 3\cdot 3}\\\\\\=10(-1\cdot 2\cdot 2\cdot 2\cdot 3)\\\\=10(-24)\\\\=\boxed{-240}$

4 0

4 years ago

Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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