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amm1812
3 years ago
12

During your cycling trip, you and your friends often stayed in hostels. The prices of these are listed below.

Mathematics
1 answer:
saw5 [17]3 years ago
4 0

Answer:

$26.3 night-time rest days

Range is $33.19

Step-by-step explanation:

We need to find the mean of the standard cost to find the night-time rest days:

Mean=\frac{\text{sum of the observations}}{\text{Number of observations}}

So, the required mean is : \frac{22.45+18.55+24.50+11.20+??}{5}=20.60

\frac{76.7+?}{5}=20.60

\Rightarrow 76.7+?=103

\Rightarrow ?=26.3

After simplification we will get the night-time rest days price which is: $26.3

Now, we will find the range in cost of stays in standard price and with breakfast price which is difference between the maximum cost minus the lowest cost

So, the sum of Standard cost is:  $103

And the sum of Breakfast price cost is: $136.19

So, the required range is: $136.19-$103=$33.19

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3 years ago
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Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

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Solve the system of equations:

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plug into 4x+4y=32 to solve for y

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Plug into x-y=2---> x-3=2---> x=5

Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

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3 years ago
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xz_007 [3.2K]

Answer:

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Step-by-step explanation:

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(x-h)^2+(y-k)^2=r^2 where (h,k) is the center and r is the radius.

So if we replace (h,k) with (0,0) since the center is the origin and r with 2 since the radius is 2 we get:

(x-0)^2+(y-0)^2=2^2

Let's simplify:

x^2  +  y^2   = 4

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3 years ago
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