researchers fit data from over 6000 fetal ultrasounds. they measured the femur length, l, (in millimeters)…

researchers fit data from over 6000 fetal ultrasounds. they measured the femur length, l, (in millimeters) as a function of the fetus age, t, (in weeks) and found the formula below. calculate the rate of growth, $\\frac{d l}{d t}$, at t = 10, 15, and 25 weeks. does the rate of growth of the fetus increase or decrease as it ages?\n\nl=-37.60 + 3.71t-6.32×10^{-4}t^{3}\n\nwrite a function for $\\frac{d l}{d t}$.\n\n$\\frac{d l}{d t}=3.71 - 0.001896t^{2}$\n\n(use integers or decimals for any numbers in the expression.)\n\ncalculate the rate of growth at t = 10 weeks.\n\n$\\frac{d l}{d t}=\\square mm/week$\n\n(round to two decimal places as needed.)

researchers fit data from over 6000 fetal ultrasounds. they measured the femur length, l, (in millimeters) as a function of the fetus age, t, (in weeks) and found the formula below. calculate the rate of growth, $\\frac{d l}{d t}$, at t = 10, 15, and 25 weeks. does the rate of growth of the fetus increase or decrease as it ages?\n\nl=-37.60 + 3.71t-6.32×10^{-4}t^{3}\n\nwrite a function for $\\frac{d l}{d t}$.\n\n$\\frac{d l}{d t}=3.71 - 0.001896t^{2}$\n\n(use integers or decimals for any numbers in the expression.)\n\ncalculate the rate of growth at t = 10 weeks.\n\n$\\frac{d l}{d t}=\\square mm/week$\n\n(round to two decimal places as needed.)

Answer

Explanation:

Step1: Substitute ( t = 10 ) into ( \frac{dL}{dt}=3.71 - 0.001896t^{2} )

[ \begin{align*} \frac{dL}{dt}&=3.71-0.001896\times(10)^{2}\ &=3.71 - 0.001896\times100\ &=3.71-0.1896 \end{align*} ]

Step2: Calculate the value

[ \frac{dL}{dt}=3.5204\approx3.52 ]

Answer:

(3.52)