Find the approximate area of the composite figure below

How do the prefixes in the metric system relate to the basic units?

WITCHER [35]

The basic unit of length in the metric system is the meter. Grams are used to measure weight and liter is used to measure liquid capacity
besides, it is used so that you can express very large and very small numbers in a short unambiguous manner.
Example Instead of saying "three thousand million" meters
They classify the decimal place of the basic unit. ie, the meter is classified with milli (1/1000), centi(1/100), deci (1/10), Deca (10), Hecto (100), Kilo (1000).

4 0

3 years ago

Read 2 more answers

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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

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

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago

There are 3 red marbles, 5 green marbles, and 2 blue marbles in a jar. Event A is defined as drawing a red marble from the jar o

Anastaziya [24]

P(B/A) is $\frac{5}{9}$ in the simplest form

Step-by-step explanation:

Let us explain how to solve the problem

P(B|A) is called the "Conditional Probability" of B given A
Conditional probability is the probability of one event occurring with some relationship to one or more other events that means event A has already happened, now what is the chance of event B
The formula for conditional probability is P(B|A) = P(A and B)/P(A), where P(A and B) is P(A) . P(B given A occurred)

∵ There are 3 red marbles, 5 green marbles, and 2 blue

marbles in the jar

∴ The total number of the marbles in the jar = 3 + 5 + 2 = 10

∵ Two marbles are drawn from the bag, one after the other without

replacement

- That means red is already occurred when green is happened,

so it is a "Conditional Probability" of green given red

P(B/A) = P(A and B)/P(A)

∵ Event A is defined as drawing a red marble from the jar on

the first draw

∴ P(A) = P(red)

∵ P(red) = $\frac{3}{10}$