How much is 19.5% of 600? Round the answer to the nearest whole number.

Alex73 [517]

Answer:

BC

Parallel: Two sides or lines are parallel if they are lines that are always the same distance from each other and will never intersect or touch

8 0

3 years ago

Read 2 more answers

Help Please Asappppppppp I need help!!!

Firlakuza [10]

Answer: 9, 1/20, 9/20

Step-by-step explanation:

1) 9 pieces

2) Size of each pieces:

$=\frac{3}{4} *\frac{1}{3}* \frac{3}{5} *\frac{1}{3}\\= \frac{1}{4} *\frac{1}{5} \\= \frac{1}{20}$

3) Area of the shaded rectangle:

$=\frac{3}{4} *\frac{3}{5} \\= \frac{9}{20}$

4 0

3 years ago

I could use some help!!

12345 [234]

Answer:

84

Step-by-step explanation:

The interquartile range is obtained using the relation :

Third quartile (Q3). - First quartile (Q1)

From. The boxplot :

Q3 = 96

Q1 = 12

Interquartile range (IQR) = Q3 - Q1 = 96 - 12 = 84

6 0

3 years ago

A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi

Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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

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

For the sum of n values of a distribution, the mean is $\mu \times n$ and the standard deviation is $\sigma\sqrt{n}$

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that $\mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300$