The most appropriate time-sampling procedure when a person must collect data and engage in another activity simultaneously is?

1 answer:

5 0

The most appropriate time-sampling procedure when a person must collect data and engage in another activity simultaneously is momentary sampling with longer interval

Time sampling is a method of sampling behavior in an observation study and is where an observer records behavior at prescribed intervals.

Momentary Time Sampling refers to recording the occurrence behavior data only if the student is engaged in that behavior at the end of the interval.

Here, a person must collect data and engage in another activity simultaneously, that is momentary sampling with longer interval

Hence, the most appropriate time-sampling procedure when a person must collect data and engage in another activity simultaneously is momentary sampling with longer interval

Learn more about Time sampling here

brainly.com/question/18630925

#SPJ4

You might be interested in

PLEASE HELP !!!!!!!! prove that segment DE is parallel to segment AB.

RideAnS [48]

Angle A is equal to Angle D because they are opposite interior angles I believe. Same thing goes for angles B and E.

3 0

3 years ago

Read 2 more answers

14-3(5t-12)=1-(20t+1)<br> How am I supposed to do this?

Nuetrik [128]

Expand by multiplying out:

14-15t + 36 = 1 -20t -1,
Because of the minus before the bracket, it changes the sign inside the bracket
-15t + 20t = 1-1-14-36 Rearranging the ts on one side and the numbers on other
side.
5t = -50 Divide both sides by 5.
t = -50/5
t = -10.

5 0

4 years ago

AleksAgata [21]

$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\textit{sphere A}}{\textit{sphere B}}\qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}\qquad \qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}=\stackrel{\stackrel{volumes'}{ratio}}{\cfrac{\sqrt[3]{288}}{\sqrt[3]{v}}}\implies \cfrac{1}{4}=\sqrt[3]{\cfrac{288}{v}}\implies \left( \cfrac{1}{4} \right)^3=\cfrac{288}{v} \\\\\\ \cfrac{1^3}{4^3}=\cfrac{288}{v}\implies \cfrac{1}{64}=\cfrac{288}{v}\implies v=18432$