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2 years ago

How to find the new interquartile range of the numbers 9,10,11,12,16,20,20

Mathematics

2 answers:

UNO [17]2 years ago

6 0

10 is ur range

hope this helps

muminat2 years ago

6 0

Answer:

The interquartile range would be 10.

Step-by-step explanation:

9, 10, 11, 12, 16, 20, 20

Median: 12

Lower quartile (Q1): 10

Upper quartie (Q2): 20

(IQR) Interquartile range: 20 - 10 = 10

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Can someone help me with this problem? thank you!

abruzzese [7]

Answer:

When 6x - 4 = 8, x = 2.

Step-by-step explanation:

6x - 4 = 8

Add 4 to both sides.

6x = 12

Divide both sides by 6.

x = 2

Proof:

6x - 4 = 8

Substitute variable.

6(2) - 4 = 8

Multiply 6 and 2.

12 - 4 = 8

Subtract 4 from 12.

8 = 8

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1 year ago

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Scilla [17]

Answer:

$\checkmark \text{B. The area of the circle is } 729\pi,\\\checkmark \text{D. The arc length of the sector is } 12\pi$

Step-by-step explanation:

Area of a circle with radius $r$ : $r^2\pi$

Circumference of a sector with radius $r$ : $2r\pi$

Area of sector with angle $\theta$ : $r^2\pi\cdot \frac{\theta}{360}$ .

Arc length of sector with angle $\theta$ : $2r\pi\cdot \frac{\theta}{360}$

Using these equations, we get the following information:

Area of circle: $27^2\pi=729\pi$

Circumference of a circle: $2\cdot 27\cdot \pi=54\pi$

Area of sector: $27^2\pi\cdot \frac{80}{360}=162\pi$

Arc length of sector: $2\cdot 27\cdot \pi \cdot \frac{80}{360}=12\pi$

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2 years ago

What place value are the 8s in 888,000?

OleMash [197]

Answer:

1,000

10,000

100,000

Step-by-step explanation:

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2 years ago

What is the range and domain of f(x)=60-0.18x

Gnom [1K]

The domain is always the value of x, but there is no value of x shown..

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2 years ago

A survey of 25 grocery stores revealed that the mean price of a gallon of milk was $2.98, with a standard error of $0.10. What i

Gnesinka [82]

Answer:

95% confidence interval: (2.784,3.176)

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 25

Sample mean = $2.98

Standard error = $0.10

Alpha = 0.05

95% confidence interval:

$\mu \pm z_{critical}(\text{Standard error})$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

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2 years ago