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Goshia
9 days ago
8

The data set represents the number of rings each person in a room is wearing.

Mathematics
2 answers:
zzz [12.3K]9 days ago
4 0

Result:

The interquartile range for the data is:

                                4

Detailed explanation:

A data set is provided as:

             0, 2, 4, 0, 2, 3, 2, 8, 6

Arranging this data in ascending order results in:

    0     0     2     2    2    3      4      6     8

The lowest value in the data set=0

Highest value in the data set is: 8

Data set's range= Highest value-Lowest value

i.e. Range= 8-0

i.e. Range= 8

The median, which represents the central value of the data, is calculated as:

Median=  2

The lower half of the data consists of:

                0    0     2    2

Hence, the median of this lower half is the lower quartile or first quartile.

Thus, Q_1

Therefore, Q_1=\dfrac{0+2}{2}\\\\\\Q_1=\dfrac{2}{2}\\\\\\Q_1=1

Thus, Lower quartile=1

Similarly, the upper half of the data includes:

                    3      4      6     8

Consequently, the median for this upper half is the upper quartile or third quartile.

Thus, Q_3

Thus, Q_3=\dfrac{4+6}{2}\\\\\\Q_3=\dfrac{10}{2}\\\\\\Q_3=5

Therefore, Upper quartile=5

Thus, the interquartile range (IQR) can be expressed as:

IQR=Upper quartile-Lower quartile

IQR=5-1

                                    IQR=4

Inessa [12.5K]9 days ago
3 0
For the given data set reflecting the number of rings each individual is wearing: 0,2,4,0,2,3,2,8,6, the interquartile range is determined to be 2. Here, Q1 equals 4, the median or Q2 is 3, and Q3 is 6. The method for calculating the interquartile range is IQR = Q1 - Q2.
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Answer:

The function decreases across all real numbers x where x < 1.5.

Step-by-step explanation:

We have

f(x)=(x-4)(x+1)

f(x)=x^2+x-4x-4

f(x)=x^2-3x-4

This represents an upward-opening vertical parabola

The vertex denotes a minimum point

The vertex is located at (1.5,-6.25)

We know that

The function is decreasing within the interval ----> (-∞, 1.5) x < 1.5

This means----> the function is continuously decreasing for all real x values under 1.5

Conversely, the function is increasing within the interval ----> (1.5, ∞) x> 1.5

Thus, for all real x values greater than 1.5, the function is increasing

Check the attached figure for further clarification

Therefore

The true statement is

The function decreases across all real numbers x where x < 1.5.

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1 month ago
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Before the last school year began, it was estimated that the average discretionary personal expenses each school year for a stud
tester [12383]

Answer:

M=2,179.29

The average exceeds the estimate

Her expenses would need to be $1,625.00

Step-by-step explanation:

1.) What is the mean of her friends' personal expenses?

The mean is defined as the average expense value, calculated as follows:

M=\frac{2,800+1,990+2,005+2,400+1,860+2,200+2,000}{7}\\M=2,179.29

2.) Is the average higher or lower than the estimate?

Given the estimate of $2,110, the average is indeed higher than this estimate.

3.)What amount of personal expenses would Ashley need to maintain a combined average of $2,110 with her friends' amounts for that school year?

M= 2,110=\frac{7*2,179.29+x}{8} \\x=2,110*8 - 15,255\\x=1,625

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25 days ago
A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150
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Answer:

Part 1) The equation is A(x)=150x-2x^2

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

P=2x+y

P=150\ m

thus

150=2x+y

y=150-2x -----> this is equation A

The area of the rectangular school yard is

A=xy ----> this is equation B

Substituting equation A into equation B yields

A=x(150-2x)

A=150x-2x^2

Change to function notation

A(x)=150x-2x^2

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

A(40)=150(40)-2(40^2)

A(40)=2,800\ m^2    

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

A(x)=150x-2x^2

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

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Answer:

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1 month ago
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