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The Lone Star Shipping Company charges $14 plus $2 a pound to ship an overnight package. Discount Shipping Company charges $20 p

Ymorist [56]

Answer:

12 pound

Step-by-step explanation:

Let $x$ be the weight of the package.

For Lone Star Shipping, the cost will be:

$2x + 14$

For Discount Shipping Company, the cost will be:

$1.5x + 20$

We hence equate these two expressions equal to each other, since the cost should be same.

$2x + 14 = 1.5x + 20 \\2x - 1.5x = 20 -14\\0.5x = 6\\x = 6 \div 0.5\\x = 12$

4 0

2 years ago

According to 2013 report from Population Reference Bureau, the mean travel time to work of workers ages 16 and older who did not

gregori [183]

Answer:

a) 48.80% probability that his travel time to work is less than 30 minutes

b) The mean is 30.7 minutes and the standard deviation is of 3.83 minutes.

c) 13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes

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