Jacob wants to know the favorite sport among 6th grade students. Which types of survey below would give Jacob valid results?

Mathematics

2 answers:

agasfer [191]2 years ago

8 0

Answer:

Step-by-step explanation:

A is no use for 6th grade students

And B and D are bias because they are surveyed at sport places

Anastasy [175]2 years ago

4 0

Answers are A, D
Because they will have people that don’t like sports and do like sports where as if you survey them in a sports shop there gonna like sports otherwise they wouldn’t be there

You might be interested in

What is the half of 57

andriy [413]

Half of 57 is 28.5. Hoped I helped. :)

4 0

3 years ago

Read 2 more answers

Please help. will mark brainliest (part two) lol

Archy [21]

Is going to be 32 ft

4 0

3 years ago

Read 2 more answers

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$