You share only one distribution, so we'll focus on that one: mean: 30; std. dev.: 4.
Draw a "standard normal curve." Draw a vertical line in the exact middle of your curve. Label this line "30." Now "one standard dev. above the mean" is 30+4=34; "two std. devs. above the mean is 30+4+4=38, or 30+8=38. "three std. devs. above the mean is 30+3(4) = 42.
Now work in the other direction. Start with the mean: 30. But now subtract the std. dev. (4) instead of adding it. You'll get 30-4=26. This is "1 std. dev. below the mean. Continue: find 2 and 3 std. devs. below the mean.
Answer:
<h3>
= -1 </h3><h3>
*
= 1.4142</h3><h3>
= π =3.14159</h3><h3>
= π² = 9.8696</h3>
Explanation:
1.) Rewrite √24 = 
Rewrite √54 =
Divide both terms by the denominator; √6 cancels. 2-3 = -1
2.) Rewrite as 
×
The 3's cancel. 10/2 = 5√2 in the numerator.
√5 × √5 = 5 in the denominator. The 5's cancel.
That leaves √2 ≈1.4142
3.) Divide the terms in the numerator by the term in the denominator.
√2's cancel. 10π/2 = 5π 8π/2 = 4π
Subtract and we are left with π = 3.14159
4.) The square roots are reciprocals. They multiply to 1
We are left with π × π = π² ≈ 9.8696
<h3>
</h3>
Answer:
x=6
Step-by-step explanation:
There is a positive association between
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