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2 years ago

If a car can drive 62 miles on 3 gallons of gas, how many gallons of gas will it use to travel 231 miles?

Mathematics

1 answer:

aksik [14]2 years ago

8 0

Answer:

11 gallons of gas is needed to travel 231 miles.

Step-by-step explanation:

We can set up a proportion for this problem.

<em>For every 62 miles, you use 3 gallons of gas. </em>

$\frac{62}{3}*\frac{231}{?}$

Now, let's cross multiply.

$\frac{693}{62}$ ≈ $11$

So, you will need about 11 gallons of gas to travel 231 miles.

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Answer:

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2 years ago

A studemt is hiring a 3-kilometer race. he runs 1 kilometer every 2 minutes. select the function that describes his distance fro

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B. f(x) = -1/2x + 3

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2 years ago

The average weight of the top 5 fish caught at a fishing tournament was 12.3 pounds. Some of the weights of the fish are shown i

UNO [17]

Answer:

14.6

Step-by-step explanation:

It is given that,

The average weight of the top 5 fish caught at a fishing tournament was 12.3 pounds. From the attached figure, the weight of 5 fish are given. We need to find the weight of Wayne S. fish.

Average = sum of terms/no of terms

Let the weight of Wayne S. is x. So,

Here the sum of terms is x+12.8+12.6+11.8+9.7 and the number of terms is 5.

$12.3=\dfrac{x+12.8+12.6+11.8+9.7}{5}\\\\61.5=x+12.8+12.6+11.8+9.7\\\\61.5=x+46.9\\\\x=14.6$

So, the weight of the heaviest fish is 14.6.

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2 years ago

Find a number halfway between 2 1/2 and 1 1/3

kvasek [131]

Answer:

23/12

Step-by-step explanation:

2 1/2 = 5/2(3/3) = 15/6

1 1/3 = 4/3(2/2) = 8/6

15/6 - 8/6 = 7/6

7/6(1/2) = 7/12

16/12 + 7/12 = 23/12

4 0

2 years ago

One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P

Darya [45]

Answer:

The differential equation for the model is

$\frac{dP}{dt}=kP(1-P)$

The model for P is

$P(t)=\frac{1}{1-0.99e^{t/447}}$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

$\frac{dP}{dt}=kP(1-P)$

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

Solving the diferential equation

$\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}$

$P(0)= \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)= \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447$

Then the model for the population infected at time t is:

$P(t)=\frac{1}{1-0.99e^{t/447}}$

Now, we can calculate t for P(t)=90,000

$P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

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3 years ago