When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.
The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.
Answer:
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Explanation:
<em>If </em><em>there </em><em>a</em><em>r</em><em>e </em><em>more </em><em>electrons </em><em>than </em><em>protons </em><em>in </em><em>a </em><em>piece </em><em>of </em><em>matter </em><em>it </em><em>will </em><em>have </em><em>a </em><em>negative</em><em> </em><em>charge </em><em>.</em><em> </em><em>i</em><em>f</em><em> </em><em>there </em><em>are </em><em>fever </em><em>it </em><em>will </em><em>have </em><em>positive</em><em> </em><em>charge </em><em>and </em><em>if </em><em>there </em><em>are </em><em>e</em><em>qual </em><em>numbers </em><em>it </em><em>will </em><em>have </em><em>neutral</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
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hope it was helpful to you.....
Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557
Answer:
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