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jarptica [38.1K]

2 years ago

15

David owns a clothes shop on Rolling Street. Which of the following is a statistical question that David can be asked?

2 answers:

Leni [432]2 years ago

4 0

Answer:The correct answer is A

Step-by-step explanation:

Because a Statistical question is a question that have a vary of answers that can be answered by collecting data. A non-statistical question is a question that can be answered with ONE answer. For an example of non-statical is how much does that brick weigh. A is correct because it has variety of answers and and is also answered by collecting data

valina [46]2 years ago

3 0

Answer:

i would say d.

Step-by-step explanation:

because depending on whether he is a competitor seller go's by the competition prices they might want to know how low is his cheapest shirt they might want to know which one is cheaper, and he would also keep track of his sales

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2 years ago

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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

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