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umka21 [38]
3 years ago
7

We have N cars on a circular one-way road; they have the same make, same model, same year and the same fuel economy. The total a

mount of gas in all cars is sufficient to make the full circle.
Prove by induction that it is always possible to find a car that can make the full circle, taking gas from other cars as it passes them.
Mathematics
1 answer:
Natasha2012 [34]3 years ago
8 0

Answer: Satisfied for n=1, n=k and n=k+1

Step-by-step explanation:

The induction procedure involves two steps

First is

Basic Step

Here we consider that for the value n=1, there is one car and it will always make the full circle.

Induction Step

Since basic step is satisfied for n=1

Now we do it for n=k+1

Now according to the statement a car makes full circle by taking gas from other cars as it passes them. This means there are cars that are there to provide fuel to the car. So we have a car that can be eliminated i.e. it gives it fuels to other car to make full circle so it is always there.

Now ,go through the statement again that the original car gets past the other car and take the gas from it to eliminate it. So now cars remain k instead of k+1 as it's fuel has been taken. Now the car that has taken the fuel can make the full circle. The gas is enough to make a circle now.

So by induction we can find a car that satisfies k+1 induction so for k number of cars, we can also find a car that makes a full circle.

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the diameters of Douglas firs grown at a Christmas tree farm are normally distributed with a mean of 4 inches and a standard dev
aksik [14]

Answer:

Proportion of the trees will have diameters between 2 and 6 inches = 0.8164

Step-by-step explanation:

Given -

Mean (\nu )  = 4

Standard deviation (\sigma  ) = 1.5

Let X be the diameter of tree

proportion of the trees will have diameters between 2 and 6 inches =

P(2<  X<  6)   =  P(\frac{2 - 4 }{1.5}< \frac{X - \nu }{\sigma}<  \frac{6 - 4 }{1.5})

                         = P(\frac{-2 }{1.5}< Z<  \frac{2 }{1.5})     Put  [Z = \frac{X - \nu }{\sigma}]

                         =  P(-1.33< Z<  1.33)

                          = (Z<  1.33) - (Z<  -1.33)

                          = .9082 - .0918

                           = 0.8164

6 0
3 years ago
Round 2.12747677748 to the nearest ten-thousandth.
cupoosta [38]

Answer:

0

Step-by-step explanation:

7 0
3 years ago
When graphing a linear inequality, when can you NOT use (0, 0) as a test point to determine which side of a boundary line to sha
TEA [102]
When the point (0,0) is on the boundary line
8 0
3 years ago
Read 2 more answers
A garden in the shape of a trapezoid has an area of 44.4 square
Leviafan [203]

Answer:

37 meters

Step-by-step explanation:

The equation for the area of a trapezoid is A=\frac{a+b}{2} +h.

Here, a = 4.3, b=10.5, and h is unknown.

44.4=\frac{4.3+10.5}{2} +h

44.4=\frac{14.8}{2} +h

44.4=7.4 +h

37 = h

The height of the trapezoid is the width of the garden, so the garden is 37 meters wide.

5 0
3 years ago
Mr Smith's art class took a bus trip to an art museum. The bus averaged 65 miles per hour on the highway and 25 miles per hour i
Leya [2.2K]
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
\left \{ {{x+y=375} \atop { \frac{1}{65}x+ \frac{1}{25}y =7}} \right.
 
Now, the system of equations in matrix form will be:
\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right]   \left[\begin{array}{ccc}x&\\y&\end{array}\right] =  \left[\begin{array}{ccc}375&\\7&\end{array}\right]

Next, we are going to find the determinant:
D=  \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}
Next, we are going to find the determinant of x:
D_{x} =  \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8

Now, we can find x:
x=  \frac{ D_{x} }{D} = \frac{8}{ \frac{8}{325} } =325mi

Now that we know the value of x, we can find y:
y=375-325=50mi

Remember that time equals distance over velocity; therefore, the time on the highway will be:
t_{h} = \frac{325}{65} =5hours
An the time on the city will be:
t_{c} = \frac{50}{25} =2hours

We can conclude that the bus was five hours on the highway and two hours in the city. 

8 0
3 years ago
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