A boat takes 10 hours to sail a distance of 100 nautical miles with the current.

Mathematics

1 answer:

Inessa [10]2 years ago

5 0

The rate of the current is 6 mile/hour.

In this context, the rate is tantamount to speed o velocity in a linear motion. The S.I unit will be mile per hour.

Given that a boat takes 10 hours to sail a distance of 100 nautical miles with the current. The rate at which the boat move will be

The rate R1 = 100/10

R1 = 10 mile/hour

And when it moves against the current, it takes the boat 25 hours to sail the same distance. The rate at which the boat move will be

The rate R2 = 100/25

R2 = 4 mile/hour

The rate of the current R will be the difference between the two calculated rate. That is,

R = R1 - R2

R = 10 - 4

R = 6 mile/hour

Therefore, the rate of the current is 6 mile/hour.

Learn more about Rate here: brainly.com/question/8728504

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The z-score for this data is Z = -0.26.

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$