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mina [271]
3 years ago
7

A marketing company wants to know if people who like brand X soda also like brand Y potato chips. Which procedure is the most ap

propriate to answer the question?
A. Randomly choose two groups. Ask one group if they like brand X soda and ask the other group if they like brand Y potato chips.

B. Randomly choose a group. Ask them if they like both brand X soda AND brand Y potato chips.

C. Ask a random group to choose two things they like from a list of 10 food items, where two of the choices are brand X soda and brand Y potato chips.

D. Ask a random group two questions: Do you like brand X soda? Do you like brand Y potato chips?
Mathematics
1 answer:
k0ka [10]3 years ago
4 0

Answer:

D. Ask a random group two questions: Do you like brand X soda? Do you like brand Y potato chips?

Step-by-step explanation:

The correct answer is D.

The group has to be chosen randomly. However, in option B, the person could simple answer no if asked if the liked both soda and potato chips and they only liked one.

In D, the person would have to answer for both, and it would be possible to estabilish dependence or independence, for example.

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Art [367]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166243

——————————

Solve the trigonometric equation:

     \mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}


Make a substitution:

     \mathsf{cos\,x=t\qquad (-1\le t\le 1)}

and the equation becomes

     \mathsf{2t^2-t-1=0}


Rewrite conveniently  – t  as  + t – 2t,  and then factor the left-hand side by grouping:

      \mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}


Factor out  2t + 1:

     \mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}


Substitute back for  t = cos x:

     \begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}


Therefore,

     \begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}

where  k  is an integer.


Solution set:   

\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}


I hope this helps. =)

3 0
3 years ago
Enter your answer and show all the steps that you use to solve this problem.
slamgirl [31]

Answer:

h = -t+20

After 8 hours the candle will be 12 inches

Step-by-step explanation:

We have 2 points (time, inches)

 (3,17)  and (5,15)

We can find the slope

m = (y2-y1)/(x2-x1)

   = (15-17)/(5-3)

   = -2/2

   = -1

We can use point slope form

y-y1 = m(x-x1)

Replace y with h and x with t

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Distribute

h-17 = -t+3

Add 17 to each side

h-17+17 = -t+3+17

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3 years ago
Given a table of values of a quadratic function, which of the following do you absolutely need in order to find its equation? *
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Answer:

The general form of a quadratic function is --> y = ax2 + bx + c

Since (0, -2) exists for the function, we can plug in those values:

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So now our function so far is --> y = ax2 + bx - 2

We have two pairs of coordinates left: (-1, -8) and (3, -8).

First, plug in the first pair and simplify as much as you can:

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Second, plug in the second pair and simplify as much as you can:

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elimination

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Using substitution:

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plug into the second equation and solve for b

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Step-by-step explanation:

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