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aalyn [17]
4 years ago
11

For several months, a restaurant manager collected data on the number of people at a table (p) and the total bill, in dollars (d

). She found that there is a positive linear association between p and d that is best modeled by the equation d = 15.7p + 2.0. What statement is true?
The model predicts that the average total bill for a table is $15.70.
The model predicts that for each additional 2 people at a table, the total bill increases by $15.70.
The model predicts that the average bill for any table with 2 people is $15.70. The model predicts that for each additional person at a table, the total bill increases by $15.70.
Mathematics
2 answers:
Leviafan [203]4 years ago
4 0
The answer will be c. I'm sorry if I'm wrong 
Vlada [557]4 years ago
4 0
Im 100% sure its C.
<span>The model predicts that the average bill for any table with 2 people is $15.70.</span>
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Eduardwww [97]

Answer:

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Step-by-step explanation:

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the length of the rectangle is twice its width. write and solve a system of linear equations to find the length L and width W of
xz_007 [3.2K]

Answer:

2 ( m + 2 m)  = 36 is the required linear equation

Length of the rectangle is 12 ft and width =  6 ft

Step-by-step explanation:

Let the width of the rectangle = m ft.

So, the length of the rectangle = 2m  ft.

Perimeter  of rectangle = 36 ft

Now, PERIMETER OF THE RECTANGLE = 2(LENGTH + WIDTH)

⇒2 ( m + 2m)  = 36

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5 0
3 years ago
The manufacturer of a CD player has found that the revenue R​ (in dollars) is Upper R (p )equals negative 5 p squared plus 1 com
AleksAgata [21]

Answer:

The maximum revenue is $1,20,125 that occurs when the unit price is $155.

Step-by-step explanation:

The revenue function is given as:

R(p) = -5p^2 + 1550p

where p is unit price in dollars.

First, we differentiate R(p) with respect to p, to get,

\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550

Equating the first derivative to zero, we get,

\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155

Again differentiation R(p), with respect to p, we get,

\dfrac{d^2(R(p))}{dp^2} = -10

At p = 155

\dfrac{d^2(R(p))}{dp^2} < 0

Thus by double derivative test, maxima occurs at p = 155 for R(p).

Thus, maximum revenue occurs when p = $155.

Maximum revenue

R(155) = -5(155)^2 + 1550(155) = 120125

Thus, maximum revenue is $120125 that occurs when the unit price is $155.

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4 years ago
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