Help its ugent ill mark brainleyest

1. Identify the DEPENDENT VARIABLE in the testable

MAVERICK [17]

To identify the dependent variable in the testable, look out for the variable that is affected by the other. The variable that changes as a result of another variable is the dependent variable.

In a research study, there are typically two main variables that direct the scientific enquiry. They are:

Dependent Variable, and
Independent Variable

The independent variable causes a change in the dependent variable, i.e. the dependent variable receives the effect, the independent variable is the cause of the change.

It is very easy to identify the dependent variable in any testable hypothesis once you are able to pick out which variable is causing a change in the other.

For example, let's say the topic of a research is: The Impact of Sunlight on Germination Rate of Seedlings.

Here, Sunlight is the independent variable affecting Germination Rate.

The dependent variable here would be: Germination Rate.

Therefore, to identify the dependent variable in the testable, look out for the variable that is affected by the other. The variable that changes as a result of another variable is the dependent variable.

Learn more here:

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5 0

2 years ago

If a=2, b=3 and c=4 then find the numerical value of a2+2abc+b2+c2

qaws [65]

a=2 b=3 and c=4. then,

a2+2abc+b2+c2

a2+2abc+b2+c2= 2^2+2×2×3×4+3^2+4^2

(replace value of a, b, c by2,3,4 respectively tgen solve)

= 4+48+9+16

= 77………………

Therefore, 77 is correct answer.

7 0

3 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$