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3 years ago

Use the drop down menus to complete the statements to match the information shown by the graph.

Mathematics

2 answers:

jenyasd209 [6]3 years ago

8 0

Answer:

less than, slope, higher (the rest in not sure)

Step-by-step explanation:

just guessing, could you please show the drop downs??

Zepler [3.9K]3 years ago

6 0

Answer:

Produce per pound costs less than meat per pound because the slope of the graph for meat is greater that of the graph for produce.

The slope the graph tells you the unit rate in dollars per pound.

Step-by-step explanation:

In the graph the x-axis represents weight in the pound and y-axis represents the price in dollar,

Since, slope of a line represents unit rate of change and it is constant.

So, slope the graph represents the unit rate in dollars per pound.

Also, the steeper line( closer to y-axis ) has higher slope.

By the given graph,

Graph of line represents cost of meat per pound is steeper than that of cost of produce per pound.

Thus, the cost of meat per pound is more than cost of produce per pound because the graph for meat is greater that of the graph for produce.

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2 years ago

An advertising company designs a campaign to introduce a new product to a metropolitan area of population 3 Million people. Let

Advocard [28]

Answer:

$P(t)=3,000,000-3,000,000e^{0.0138t}$

Step-by-step explanation:

Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have

$P'(t)=K(3,000,000-P(t))$

Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising

We have and ordinary differential equation of first order that we can write

$P'(t)+KP(t)= 3,000,000K$

The <em>integrating factor </em>is

$e^{Kt}$

Multiplying both sides of the equation by the integrating factor

$e^{Kt}P'(t)+e^{Kt}KP(t)= e^{Kt}3,000,000*K$

Hence

$(e^{Kt}P(t))'=3,000,000Ke^{Kt}$

Integrating both sides

$e^{Kt}P(t)=3,000,000K \int e^{Kt}dt +C$

$e^{Kt}P(t)=3,000,000K(\frac{e^{Kt}}{K})+C$

$P(t)=3,000,000+Ce^{-Kt}$

But P(0) = 0, so C = -3,000,000

and P(50) = 1,500,000

$e^{-50K}=\frac{1}{2}\Rightarrow K=-\frac{log(0.5)}{50}=0.0138$

And the equation that models the number of people (in millions) who become aware of the product by time t is

$P(t)=3,000,000-3,000,000e^{0.0138t}$

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3 years ago