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Dahasolnce [82]
3 years ago
5

Question 2 -

Mathematics
1 answer:
atroni [7]3 years ago
3 0

Coefficient of variation, CV. for Data set A is 13.5% and for Data set B is 19.6%

Sample A: 2.94, 3.06, 3.07, 3.16, 3.34, 3.43, 3.47, 3.56, 3.57, 3.97, 3.99, 4.24, 4.30, 4.35

Mean = 3.6036, Standard deviation = 0.4858.

C.V = \frac{S.D}{Mean}

=\frac{0.4858}{3.6036}

=0.1348

=0.1

Sample B: $24,800, $30,000, $22,300, $20,400, $19,000, $32,200, $23,000, $23,000, $24,000, $27,200, $34,900

Mean = 25527.2727, Standard deviation = 5001.218.

C.V = \frac{S.D}{Mean}

=\frac{5001.218}{25527.2727}

=0.1959

=0.2

In terms of percentage, C.V for sample A is 13.5% and for sample B is 19.6%

Learn more here:

brainly.com/question/12974947?referrer=searchResults

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Answer:

V≈339.29cm³

Step-by-step explanation:

4 0
3 years ago
The probability of an outcome that lies within 95% of the mean is a good indicator that it lies in which standard deviation?
kumpel [21]

Answer:

answer b

Step-by-step explanation:

Answer b is definitely the correct one.

The Empirical Rule states that 68% of normally distributed data lies within 1 standard deviation of the mean; 95% within 2 standard deviations, and 99.9% within 3.

8 0
3 years ago
The tigers won 12 games last year.this year the tigers won 20 games. What is the percent increase in the number of games the tig
sergejj [24]
Its is 60% because 12/20 is 60%

4 0
4 years ago
Please verify the following trigonometric identities.
Nina [5.8K]

Seems like there is a correction in the first question (RHS is tanx.tany)

(i) For convenience: let tanx = a ; tany = b

Thus, tanx + tany = a + b

Moreover, cotx = 1/tanx = 1/a ; coty = 1/b

Thus,

cotx + coty = 1/a + 1/b = (a + b)/ab

Hence,

=> (tanx + tany)/(cotx + coty)

=> (a + b) / { (a + b)/ab }

=> ab(a + b)/(a + b)

=> ab => tanx.tany , proved.

(ii) For convenience: let sinx = a ; cosx = b

As we know, sin²x + cos²x = 1 => a²+b²=1

=> (a³ + b³)/(a + b)

=> (a + b)(a² + b² - ab) / (a + b)

=> (a² + b² - ab)

=> 1 - ab => 1 - sinx.cosx , proved

(iii): let x/2 = A

=> tan(x/2) + cosx.tan(x/2)

=> tanA + cos2A.tanA

=> tanA [1 + cos2A]

=> tanA (2cos²A) {1+cos2A = 2cos²A}

=> (sinA/cosA) (2cos²A)

=> sinA (2cosA)

=> 2sinAcosA

=> sin2A

=> sin2(x/2)

=> sinx proved

Letting x/2 = A is not mandatory. I did it to decease words*(in a line).

<u>Indentities used</u>:

• sin²A + cos²A = 1

• (a³ + b³) = (a + b)(a² + b² - 1)

• 1 + cosA = 2 cos²(A/2)

• tanA = sinA/cosA.

• 2sinAcosA = sin2A

8 0
3 years ago
Read 2 more answers
A freight train takes 18 hours to travel the same distance that an express train travels in 15 hours. The rate of the express tr
Eva8 [605]
<span>The rate of the express train is 15 mph faster than the freight train.
Let's grasp onto this piece of information. 

The rate at which the freight train goes will be labeled "x" (keep in mind rate is just the speed)

If the freight train has a speed of "x" then the express train which is going 15mph faster than the freight train can be represented as "x + 15mph."

Our result is in HOURS, but our speed is in Miles per Hour (Miles/Hour). Meaning we need to multiply our speeds by some time (t).

Let's set up two equations: 
Freight Train: 18 hours * x mph = distance
Express Train: 15 hours * (x mph + 15mph) = distance

They are both the same distance so we can set the two equations equal to each other like this: 

18 hours * x mph = 15 hours * (x mph + 15mph)

That right side is a bit of a thorn in our side, so let's simplify it by multiplying all the terms in the parenthesis by 15 hours (keep in mind the hours in the units cancel out). 

18 hours * x mph = 15x miles + 15 * 15 miles
18 hours * x mph = 15x miles + 225 miles

Let's simplify the left side (</span>\frac{miles}{hour}*hour = miles)

18x miles = 15x miles + 225 miles

Get all the "x" terms on one side by subtracting 15x on both sides. 

18x - 15x = 225 miles

Simplify

3x = 225 miles

Divide by 3 on both sides to solve for x. 
225/3 

225 = 25*9

(25*9)/3 = 25*3 = 75mph

So "x" which we chose was the rate of the freight train is 75mph. 
Adding 15mph to the freight train rate gives you the express trains rate which is 90mph. 

Hope that helped. 
8 0
3 years ago
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