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OleMash [197]
3 years ago
8

A vendor converts the weights on the packages she sends out from pounds to kilograms (1kg approximately equals 2.2 pounds).

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

Answer:

a) b) the standard deviation and the mean is affected by the conversion factor as well

c)  the mean is displaced by b units

Step-by-step explanation:

for a new variable

Y=a*X  , where a= constant (conversion factor= 1 kg/2.2 pounds)

then

p(y)= p(a*X) = p(X)

a) mean =μ=E(Y)= ∑ a*X*p(y) = a ∑ X*p(x) = a* E(X)

mean =μ=a*μₓ

b) σ² = ∑ (Y-μ)²* p(y) =  ∑ (a*X-a μₓ)²* p(y) = a²*∑ (X-μₓ)²* p(x) = a²*σₓ²

then

standard deviation = σ= √σ²=√(a²*σₓ²) = a*σₓ

standard deviation = σ= a*σₓ

then the standard deviation and the mean is affected by the conversion factor as well

c) nevertheless for a displacement b

Y₂=X + b (b= constant= 50 gr)

p(Y₂)= p(X + b) = p(X)

then

mean =μ=∑ (X-b)*p(y)=∑ X*p(x)- b ∑ p(x) = E(X) -

mean =μ=μₓ - b

then the mean is displaced by b units

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Answer:

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Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

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