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GarryVolchara [31]
1 year ago
10

The sets of data below show the heights, in inches, of students in two different preschool classes. A box plot titled Class 1. T

he number line goes from 38 to 49. The whiskers range from 39 to 48, and the box ranges from 40 to 43. A line divides the box at 41. Class 1 A box plot titled Class 2. The number line goes from 38 to 49. The whiskers range from 38 to 49, and the box ranges from 39 to 42. A line divides the box at 41. Class 2 The teachers of the two classes want to compare the heights of their students. Which statements about the data sets are accurate? Select three options. Because the sets are symmetrical, the mean should be used to compare the data sets. 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. Because the sets contain outliers, the median should be used to compare the data sets. The mean and mode cannot be accurately determined based on the type of data display.
Mathematics
2 answers:
lyudmila [28]1 year ago
8 0

Answer:

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

• Because the sets contain outliers, the median 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:

When we observe the set of given data above, we can denote that the data obtained by comparing the height of students from class 1 and class 2 would not be similar hence we can say this obtained data is not symmetrical.

Due to the fact that this data is is obtained from different classes it is certain that there would be variations in the data when measuring the heights of the students and an error may occurs. These variations are referred to as OUTLIERS.

Therefore, Median or Interquartile range is the appropriate measure to be used for comparing the data sets.

lesya692 [45]1 year ago
4 0

Answer:

c, d, and e are correct

Step-by-step explanation:

just trust me

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A home improvement store sold wind chimes for $30 each. A customer signed up for a free membership card and received a 5% discou
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the final price of the wind chime is $29.925 .

<u>Step-by-step explanation:</u>

Here we have , A home improvement store sold wind chimes for $30 each. A customer signed up for a free membership card and received a 5% discount off the price. Sales tax of 5% was applied after the discount. We need to find What was the final price of the wind chime . Let's find out:

Initially we have , 5% discount in $30 i.e.

⇒ (30(5))/100

⇒ (150)/100

⇒ $1.5 , So now price becomes $30 - $1.5 = $28.5 . At this amount there's 5% tax i.e.

⇒ (28.5(5))/100

⇒ (142.5)/100

⇒ $1.425

So now price becomes $28.5 + $1.425 = $29.925 . Therefore , the final price of the wind chime is $29.925 .

3 0
2 years ago
Now evaluate f(x) = 2x4 - 4x3-11x2+3x-6 for x=-2 f (-2) =
bija089 [108]
f(x)=2x^4-4x^3-11x^2+3x-6\\\\f(-2)=2\cdot(-2)^4-4\cdot(-2)^3-11\cdot(-2)^2+3\cdot(-2)-6\\\\f(-2)=2\cdot16-4\cdot(-8)-11\cdot4+3\cdot(-2)-6\\\\f(-2)=32+32-44-6-6=8
4 0
2 years ago
Read 2 more answers
A normal distribution curve, where x = 70 and σ = 15, was created by a teacher using her students’ grades. What information abou
mash [69]

Answer:

The median and mode of the students grade is 70.

Most of the students scored between 40 and 100.

Step-by-step explanation:

From the provided information it can be seen that the mean of the distribution is, <em>μ</em> = 70 and the standard deviation is, <em>σ</em> = 15.

For a Normal distributed data the mean, median and mode are the same.

So, the median and mode of the students grade is 70.

The standard deviation of the data represents the spread of the observation, i.e. how dispersed the values are along the curve.

In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68.27%, 95.45% and 99.73% of the values of a Normally distributed data lie within one, two and three standard deviations of the mean, respectively.

P(\mu-\sigma  

P(\mu-2\sigma

P(\mu-3\sigma

Assuming that maximum marks of the exam is 100, it can be said that most of the students scored between 40 and 100.

3 0
2 years ago
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Alannah has two lengths of ribbon.
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Answer:

Longest possible length for each of the shorter lengths of ribbon is 9 cm because greatest common factor for both 36 and 45 is 9.

Step-by-step explanation:

Alannah has two ribbons one length is 36cm and other is 45cm.

It asked to find shorter length of ribbons that each cut into equal pieces with out no ribbon left over.

So, let's find greatest common factor for both 36 and 45.

Let's prime factor each number

36= 2*2*3*3

45= 3*3*5

So, GCF is product of common factors for both numbers.

GCF= 3*3 =9

So, longest possible length for each of the shorter lengths of ribbon is 9 cm.

Learn more about GCF in brainly.com/question/21612147.

7 0
1 year ago
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Fittoniya [83]

Answer:

140

Step-by-step explanation:

you multiply the amount of people by the amount of floors, 7x20=140

7 0
2 years ago
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