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Katen [24]
3 years ago
13

The population standard deviation of the numbers 3, 8, 12, 17, and 25 is 7.563 correct to 3 decimal places. what happens if each

of the five numbers is multiplied by 3
Mathematics
1 answer:
user100 [1]3 years ago
7 0
Let's figure this out as though we have no idea what the answer would be.

Step One
Find the new five numbers.
3*3, 8*3, 12*3, 17*3, 25*3
9 , 24 , 36, 51, 75

Step 2
Find the average
(9 + 24 + 36 + 51 + 75)/5 = 195/5 = 39

Step 3
Subtract the individual numbers from the average
(39 - 9) = 30
(39 -24) = 15
(39 - 36) = 3
(39 - 51) = - 12
(39 - 75) = -36

Step 4 
Square the results from Step 3
30^2 = 900
15^2 = 225
3^2 = 9
(-12)^2 = 144
(-36)^2 = 1296

Step 5
Take the average of the results from step 4
(900 + 225 + 9 + 144 + 1296)/5
2574 / 5 = 514.8

Step 6
Take the square root of the result from step 5
deviation = sqrt(514.8)
deviation = 22.689

Step seven 
Compare the two standard deviations.
s2/s1 = 22.689 / 7.563 = 3

Conclusion
If you are given 1 set of numbers to find a population standard deviation and you multiply each member by a, then the result will be a * the standard population deviation of the first set of numbers.

Note
Your calculator will do this as well, but you have to know how to enter the data into your calculator. That requires that you follow the directions carefully. 

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VladimirAG [237]

Answer:

Emily: 13, Peter: 9, Joshua: 18

Step-by-step explanation:

Emily = x

Peter = Emily - 4 = x - 4

Joshua = 2(Peter) = 2(x - 4)

Total = 40 = Emily + Peter + Joshua = (x) + (x - 4) + 2(x - 4)

40 = 2x - 4 + 2x -8

40 = 4x - 4 - 8

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2 years ago
which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater
Nutka1998 [239]

Answer:

\frac{\sqrt[3]{16y^4}}{x^2}

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices \sqrt[n]{a}  = a^{\frac{1}{n}}

So,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } gives

\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}

Also from laws of indices

(ab)^n = a^nb^n

So, the above expression can be further simplified to

\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}

Multiply the exponents gives

\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

Substitute 2^5 for 32

\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

From laws of indices

\frac{a^m}{a^n} = a^{m-n}

This law can be applied to the expression above;

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})} becomes

2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}

Solve exponents

2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}

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From laws of indices,

a^{-n} = \frac{1}{a^n}; So,

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}} gives

\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}

The expression at the numerator can be combined to give

\frac{(2y)^{\frac{4}{3}}}{x^2}

Lastly, From laws of indices,

a^{\frac{m}{n} = \sqrt[n]{a^m}; So,

\frac{(2y)^{\frac{4}{3}}}{x^2} becomes

\frac{\sqrt[3]{(2y)}^{4}}{x^2}

\frac{\sqrt[3]{16y^4}}{x^2}

Hence,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } is equivalent to \frac{\sqrt[3]{16y^4}}{x^2}

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What does 6 more than twice mean?
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Answer:

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Step-by-step explanation:

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Suppose that a tunnel is dug through the center of the earth. You stand on a tower 100 m tall and drop a 3-kg rock into the tunn
Mashutka [201]

Answer:

The frequency of the resulting harmonic motion is 0.000219 Hz

Step-by-step explanation:

We are going to calculate the time it takes for one single wave ocillation.

Frequency and the time taken to finish a single wave oscillation are inversely proportional. The formula for calculating frequency when given the time taken to complete a wave cycle is written as: f = 1 / T

In this formula, f represents frequency and T represents the time period or amount of time required to complete a single wave oscillation.

I consider the initial speed to be zero, because it is of no significance compared with the free fall into the earth, through the earth and back again.

Given from Wikipedia:

The diameter of the earth is 1.2742 * 10⁴ km which is 1.27 * 10⁷ m

2 times the radius = diameter, so the radius of the earth = (1.27 * 10⁷ m) /2 = 6.4 * 10⁶ m

radius earth = r

r = 6.4 * 10⁶ m

Now imagine the tunnel and the free fall.

1. Initially the rock has no speed.

2. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

3. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed and it has travelled the distance r !

4. After this moment, the Rock will be slowed down because of the negative accelleration...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely, and again passing the distance r.

5. Now at the other end of the earth there is the same initial situation as described at point 1, only the Rock has travelled the distance equal to the diameter of the earth, (exactly 2 times r).

So basically, the samething happens once more, only this time it starts exactly from the other end of the earth...

6. Initially the rock has no speed.

7. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

8. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed.

9. By now the Rock will be slowed down because of the negative accelleration... It is moving towards the initial starting point...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely.

10. Now finally the Rock is exactly at the starting position.

In reality there will have been some loss of speed due to friction, so the Rock will be slightly lower then the 100 m above the ground.

let's calculate the time it takes to free fall for the distance r.

initial speed =0 and after 6.4 * 10⁶ m it's speed will be maximum. We need to find out how much time passes before that distance is passed.

r = v*t + 0.5*a*t²

r = 0 + 0.5*a*t²

0.5*a*t² = r

t² = r / ( 0.5 * a )

t² = 6.4 *10⁶ / ( 0.5 * 9.8 )

t² = 1.306 * 10 ⁶

t = 1142.86 s

Now please confirm that in order for the Rock to move back to the initial starting point it has to travel 4 times as much time. It has to travel r to centre of the earth then another r to travel to to the other side of the earth, and back again. So indeed 4 times r.

The time it will take must be the same as 4 * 1142.86 s

now this is the time of one single wave ocillation.

Since T = 4571.43 s

f = 1 / 4571.43

f = 0.00021874993164 Hz

The frequency of the resulting harmonic motion is 2.19 *10-4

The frequency of the resulting harmonic motion is 0.000219 Hz

6 0
3 years ago
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