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3 years ago

The sets of data below show the heights, in inches, of students in two different preschool classes.

Mathematics

1 answer:

MAXImum [283]3 years ago

6 0

Answer:

Because the sets do not contain outliers, the MAD should be used to compare the data sets.

Because the sets are not symmetrical, the IQR should be used to compare the data sets.

The mean and mode cannot be accurately determined based on the type of data display.

Step-by-step explanation:

Outliers are points besides whisker limits

In symmetrical box plots, distance between Q1 and Q2 is equal to distance between Q2 and Q3

Box plots don't show mode and mean

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6x+8y=5 and 3x+4y=-2

Feliz [49]

ANSWER If you solve for x you get -4/3 y+5/6 If you solve for y you get -3/4x +5/8.

Step-by-step explanation

and then the second answer is

If you solve for x you get-4/3y +-1/2

If you solve for y you get -3/4x +-1/2

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3 years ago

Delta Math:<br> Finding Angles (Level 1) 1/5<br> Full explanation please<br> will give brainlist

olga nikolaevna [1]

Answer:

$m\angle FGC =101\degree$

Step-by-step explanation:

$m\angle CBF = 46\degree$ (Given)

$m\angle FIG = 55\degree$ (Given)

$m\angle IFG = m\angle CBF$ (Corresponding angles)

$\implies m\angle IFG = 46\degree$

$m\angle FGC =m\angle IFG+ m\angle FIG$

(By exterior angle theorem)

$m\angle FGC =46\degree+ 55\degree$

$m\angle FGC =101\degree$

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2 years ago

A) Solve t + t + t = 15

soldi70 [24.7K]

Answer:

3t=15

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3 years ago

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The heights of adult males in the United States are approximately normally distributed. The mean height is 70 inches (5 feet 10

steposvetlana [31]

Using the normal distribution, it is found that the probability is 0.16.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

In this problem, the mean and the standard deviation are given by, respectively, $\mu = 70, \sigma = 3$ .

The proportion of students between 45 and 67 inches is the p-value of Z when <u>X = 67 subtracted by the p-value of Z when X = 45</u>, hence:

X = 67:

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

$Z = \frac{67 - 70}{3}$

Z = -1

Z = -1 has a p-value of 0.16.

X = 45:

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

$Z = \frac{45 - 70}{3}$

Z = -8.3

Z = -8.3 has a p-value of 0.

0.16 - 0 = 0.16

The probability is 0.16.

More can be learned about the normal distribution at brainly.com/question/24663213

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2 years ago