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BartSMP [9]
3 years ago
6

There is a linear relationship between a car’s weight and its gas mileage. On average, a car that weighs 2400 lb gets 35 mi/gal

and a 3000 lb car gets 29 mi/gal. Let w = car weight and g = miles per gallon. What is an equation that models the relationship between the variables?
Mathematics
1 answer:
Semmy [17]3 years ago
7 0
The required equation is
w - 2400 = (3000 - 2400)/(29 - 35) (g - 35)
w - 2400 = -100(g - 35)
w - 2400 = -100g + 3500
w + 100g = 3500 + 2400 = 5900
w + 100g = 5900
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marin [14]

Answer:

It will take 13.5 minutes for the liquid to reach - 10.8°C.

Step-by-step explanation:

The initial temperature of the experiment is 0°C and as the experiment goes on, the temperature of the liquid is changing at a rate of - 0.8°C per minute.

So, the equation that models the situation is  

T = 0 - 0.8t

⇒ T = - 0.8t ................. (1)

Where T is the temperature of the liquid in °C and t is the time of spending after the experiment starts in minutes.

Now, given T = - 10.8°C and we have to find the corresponding time (t) in minutes.

So, from equation (1) we get,  

- 10.8 = - 0.8t

⇒ t = 13.5 minutes

Therefore, it will take 13.5 minutes for the liquid to reach - 10.8°C. (Answer)

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3 years ago
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Alona [7]

Step-by-step explanation:

add all the freqency up which is 100

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3 years ago
Could someone please help me answer this question? I would really appreciate a through explanation, how to work it, the answer t
Shtirlitz [24]
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3 years ago
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tigry1 [53]
Hello, the answer in this question is choice C. ( when I simplified both of them they had the same answer which is : 10^4y-8y^3 + 3y)
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2 years ago
When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w
Lady bird [3.3K]

Answer:

a. y=6(1.7472)^x

b. y=6e^{0.558t}

c.13.3 months

Step-by-step explanation:

a.-Given the first term  at t_0 is 6 and the second term at t_3 is 32.

-Let's take rabbit population as a function of time to be

y=ab^x

where y is the population at time x and a the initial population at t_0\\

#We substitute our values to calculate the value of the constant b:

y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472

#Replace b in the population function:

y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x

Hence, the regression for the rabbit population as a function of time x is y=6(1.7472)^x

b. The exponential function in terms of base e is usually expressed as:

A=A_0e^{kt}

Where:

A_0-is the initial population at t_o

A-is the population at time t.

k-is the  exponential growth constant.

e- the exponent

Our function in terms of base exponent is rewritten as:

y=A_0e^{kt}

#Substitute with actual figures to solve for t:

y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580

Hence, the regression equation in terms of base e is y=6e^{0.558t}

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that y=6e^{0.558t}.

Therefore we calculate t as(take y=10001):

y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951

Hence, it takes approximately 13.3 months for the population to exceed 10000

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4 years ago
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