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Mathematics

1 answer:

pav-90 [236]2 years ago

5 0

Answer:

Step-by-step explanation:

<h3>Mean, Median,mode</h3>

$\boxed{Mean = \dfrac{Sum \ of \ all \ data}{number \ of data}}$

$\sf = \dfrac{234 + 247 + 236+212+248+243+176+195}{8}\\\\= \dfrac{1791}{8}\\\\= 223.875$

<h3>Median:</h3>

Arrange the numbers in ascending order.

176 , 195 , 212 , 234 , 236 , 243 , 247, 248

Number of data is even. So, median will be the average of 234 and 236

Median = average of the middle two terms.

$\sf = \dfrac{234+236}{2}\\\\ =\dfrac{470}{2}\\\\= 235$

Mode is the data which is occurring more number of times.

Range = maximum - minimum

= 248 - 176

= 72

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Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2

Mekhanik [1.2K]

Answer:

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

$y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\$

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t = e^−2(0) = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

$\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt} = -2te^{-2t} = -2(0)e^{-2(0)} = 0$