HELP PLS which box has the largest IQR

As James bought his textbooks for classes one semester, he estimated the cost to the nearest ten dollars. He knew he could cover

TEA [102]

Answer:

$310

Step-by-step explanation:

The first step is to add the costs of the textbook together :

$68.41 + $105.35 + $72.49 + $59.91 = $306.16

In order to round off to the nearest ten dollars, look at the units figure, if the number is greater or equal to 5, add 1 to the ten figure. If this is not the case, add zero. Replace the unit digit with zero

The unit digit is greater than 6, so 1 is added to the tens digit. The amount becomes $310

8 0

3 years ago

Find the sum of the first four terms of the geometric series 100 + 50 + 25 + ...

elixir [45]

Answer:

Decimal:

100 + 50 + 25 + 12.5 + 6.25 + 3.125 + 1.5625

Fraction:

$100 +50+25 + 12\frac{1}{2}+6\frac{1}{4}+3\frac{1}{8}+1\frac{9}{16}$

3 0

3 years ago

In a given year, the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500. If a samp

nika2105 [10]

Answer:

15.15% probability that the sample mean will be $192,000 or more.

Step-by-step explanation:

To solve this problem, 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$