Mathematics, 23.07.2019 02:10 becddjr

Find dy/dx and d2y/dx2, and find the slope and concavity (if possible) at the given value of the parameter. (if an answer does not exist, enter dne.) parametric equations point x = t , y = 7t − 2 t = 9

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Answer from: pebbles3961

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}$

Then for all the first derivative has a value of 7.

By the product rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\right]=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}\dfrac{\mathrm dt}{\mathrm dx}+\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm d^2t}{\mathrm dx^2}$

but $t=x\implies\dfrac{\mathrm dt}{\mathrm dx}=1\implies\dfrac{\mathrm d^2t}{\mathrm dx^2}=0$ , so we're left with

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\mathrm d^2y}{\mathrm dt^2}$

but $y=7t-2\implies\dfrac{\mathrm dy}{\mathrm dt}=7\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=0$ so the second derivative is 0 for all .

Answer from: Quest

answer: a. 13/44

step-by-step explanation: judging by the information given to us, we are looking for experimental probability. we can see that 13 males have chosen chocolate ice cream. there are 44 ice creams in total (remember that the total is always the denominator). so our answer is 13/44.

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

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