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1 year ago

Which statements about the box plot are correct? Check all that apply.

Mathematics

2 answers:

love history [14]1 year ago

6 0

Answer:

The correct options are 1 and 3.

Step-by-step explanation:

From the given box plot it is clear that

Q₁ is 25% of a data.

Median is 50% of a data.

Q₃ is 75% of a data.

34 is minimum value of the data and 46 is median it means 50% of the data values lies between 34 and 46. Therefore option 1 is correct.

42 is first quartile and and 70 is third quartile. it means 50% of the data values lies between 42 and 70. Therefore option 2 is incorrect.

The difference between Minimum value and first quartile, Maximum value and third quartile is less than 1.5×(IQR), therefore it is unlikely to have any outliers in the data.

Hence option 3 is correct.

The interquartile range of the data is

The interquartile range is 28. Therefore option 4 is incorrect.

Range of the data is

The range is 42. Therefore option 5 is incorrect.

Step-by-step explanation:

Rudiy271 year ago

3 0

Answer:

The correct options are 1 and 3.

Step-by-step explanation:

From the given box plot it is clear that

$\text{Minimum values}=34$

$Q_1=42$

Q₁ is 25% of a data.

$Median=46$

Median is 50% of a data.

$Q_3=70$

Q₃ is 75% of a data.

$\text{Maximum values}=76$

34 is minimum value of the data and 46 is median it means 50% of the data values lies between 34 and 46. Therefore option 1 is correct.

42 is first quartile and and 70 is third quartile. it means 50% of the data values lies between 42 and 70. Therefore option 2 is incorrect.

The difference between Minimum value and first quartile, Maximum value and third quartile is less than 1.5×(IQR), therefore it is unlikely to have any outliers in the data.

Hence option 3 is correct.

The interquartile range of the data is

$IQR=Q_3-Q_1$

$IQR=70-42=28$

The interquartile range is 28. Therefore option 4 is incorrect.

Range of the data is

$Range=Maximum-Minimum$

$Range=76-34=42$

The range is 42. Therefore option 5 is incorrect.

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A manager wants to build 3-sigma x-bar control limits for a process. The target value for the mean of the process is 10 units, a

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Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

Mean = 10

Sample size = 9

Standard deviation = 6

We need to find the upper and lower control limits.

so, Lower limit would be

$\bar{x}-3\dfrac{\sigma}{\sqrt{n}}\\\\=10-3\times \dfrac{6}{\sqrt{9}}\\\\=10-6\\\\=4$

Upper limit would be

$\bar{x}+3\dfrac{\sigma}{\sqrt{n}}\\\\=10+3\times \dfrac{6}{\sqrt{9}}\\\\=10+6\\\\=16$

Hence, Third option is correct.

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2 years ago

If x varies jointly as y and z, and x = 8 when y = 4 and z = 9, find z when x = 16 and y = 6. 3

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Answer:

Joint variation says that:

if $x \propto y$ and $x \propto z$

then the equation is in the form of:

$x = kyz$ , where, k is the constant of variation.

As per the statement:

If x varies jointly as y and z

then by definition we have;

$x=k(yz)$ ......[1]

Solve for k;

when x = 8 , y=4 and z=9

then

Substitute these in [1] we have;

$8=k(4 \cdot 9)$

⇒ $8 = 36k$

Divide both sides by 36 we have;

$\frac{8}{3}=k$

Simplify:

$k = \frac{2}{9}$

⇒ $x = \frac{2}{9}yz$

to find z when x = 16 and y = 6

Substitute these value we have;

$16 = \frac{2}{9} \cdot 6 \cdot z$

⇒ $16 = \frac{12}{9}z$

Multiply both sides by 9 we have;

$144 = 12z$

Divide both sides by 12 we have;

12 = z

z = 12

Therefore, the value of z is, 12

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1 year ago

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Mike watches a caterpillar climbing on a tree trunk he writes the expression -5/8-(-1/8-1/2) to show the change in feet of the c

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1 year ago

Determine whether a triangle can be formed with the given side lengths. If so, use Heron's formula to find the area of the trian

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Answer:

Triangle inequality rule states that the sum of eachpair of sides must be greater than the third sides.

240 + 123 > 207 true

240 + 207 > 123 true

123 + 207 > 240 true

A = √(s(s-a)(s-b)(s-c))

s is the semi-perimeter (a + b + c)/2

s = (240 + 123 + 207)/2

s = 285

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Step-by-step explanation:

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1 year ago

Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

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Answer: The required inequality is $x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.$ and its solution is $x>8.$

Step-by-step explanation: Given that Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

Also, Heloise has written $\frac{1}{4}$ as many articles as Mustafa has and Gia has written $\frac{3}{2}$ as many articles as Mustafa has.

We are to write an inequality to determine the number of articles, m, Mustafa could have written for the school newspaper. Also, to solve the inequality.

Since m denotes the number of articles that Mustafa could have written. Then, according to the given information, we have

$x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.$

And the solution of the above inequality is as follows :

$x+\dfrac{1}{4}x+\dfrac{3}{2}x>22\\\\\\\Rightarrow \dfrac{4x+x+6x}{4}>22\\\\\\\Rightarrow 11x>88\\\\\Rightarrow x>\dfrac{88}{11}\\\\\Rightarrow x>8.$

Thus, the required inequality is $x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.$ and its solution is $x>8.$

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