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3 years ago

13

How could two data sets could have the same means and modes but still differ greatly? Give an example with two data sets to illu

strate.

1 answer:

wariber [46]3 years ago

4 0

The sets would have different sizes and standard deviations. Examples will vary. A general method for constructing

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A flowerpot has a circular base with a diameter of 27 centimeters. Find the circumference of the base of the flowerpot. Round to

ValentinkaMS [17]

Answer:

84.82cm :}

Step-by-step explanation:

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3 years ago

No links need help ASAP

Alexandra [31]

Answer:

4, 2

Step-by-step explanation:

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3 years ago

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the ratio of boys to girls at the beach cleanup was 7:8. If there were 43 boys, how many girls were there?

faltersainse [42]

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3 years ago

PLEASE HELP ASAP ILL GIVE BRAINLIEST!!!!!<br><br>find measure of arc MK

Gwar [14]

Answer:

Arc length MK = 15.45 units (nearest hundredth)

Arc measure = 58.24°

Step-by-step explanation:

Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

$\sf \cos(\theta)=\dfrac{A}{H}$

where:

$\theta$ is the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Given:

$\theta$ = ∠KLM
A = LN = 8
H = KL = 15.2

$\implies \sf \cos(KLM)=\dfrac{8}{15.2}$

$\implies \sf \angle KLM=\cos^{-1}\left(\dfrac{8}{15.2}\right)$

$\implies \sf \angle KLM=58.24313614^{\circ}$

Therefore, the measure of arc MK = 58.24° (nearest hundredth)

$\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right) \quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle)}$

Given:

r = 15.2
∠KLM = 58.24313614°

$\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)$

$\implies \textsf{Arc length MK}=\sf 15.45132428\:units$

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2 years ago

A randomly generated list of integers from 0 to 4 is being used to simulate an event, with the numbers 1, 2, and 3 representing

Tresset [83]

Answer:

I think the anwser is

<h2>C.60 percent</h2>

hope it helped

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3 years ago