14) Since f=5, f + 69 = 74
15) Since w = 23, w + 45 = 68
Answer:
Step-by-step explanation:
In the two independent samples application, it involves the test of hypothesis that is the difference in population means, μ1 - μ2. The null hypothesis is always that there is no difference between groups with respect to means.
Null hypothesis: ∪₁ = ∪₂. where ∪₁ represent the mean of sample 1 and ∪₂ represent the mean of sample 2.
A researcher can hypothesize that the first mean is larger than the second (H1: μ1 > μ2 ), that the first mean is smaller than the second (H1: μ1 < μ2 ), or that the means are different (H1: μ1 ≠ μ2 ). These ae the alternative hypothesis.
Thus for the z test:
if n₁ > 30 and n₂ > 30
z = X₁ - X₂ / {Sp[√(1/n₁ + 1/n₂)]}
where Sp is √{ [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁+n₂-2)}
Answer:
24 cm
Step-by-step explanation:
The volume of a sphere is given by
V = 4/3 pi r^3
2304 pi = 4/3 pi r^3
Divide each side by pi
2304 = 4/3 r^3
Multiply each side 3/4
2304 * 3/4 = 4/3 r^3 * 3/4
1728 = r^3
Take the cube root of each side
1728 ^ 1/3 = r^3 ^ 1/3
12 = r
We want the diameter
d = 2*r
d = 2*12
d = 24
Based on the definition of <em>composite</em> figure, the area of the <em>composite</em> figure ABC formed by a semicircle and <em>right</em> triangle is approximately 32.137 square centimeters.
<h3>How to find the area of the composite figure</h3>
The area of the <em>composite</em> figure is the sum of two areas, the area of a semicircle and the area of a <em>right</em> triangle. The formula for the area of the composite figure is described below:
A = (1/2) · AB · BC + (π/8) · BC² (1)
If we know that AB = 6 cm and BC = 6 cm, then the area of the composite figure is:
A = (1/2) · (6 cm)² + (π/8) · (6 cm)²
A ≈ 32.137 cm²
Based on the definition of <em>composite</em> figure, the area of the <em>composite</em> figure ABC formed by a semicircle and <em>right</em> triangle is approximately 32.137 square centimeters.
To learn more on composite figures: brainly.com/question/1284145
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Answer:
μ ≈ 2.33
σ ≈ 1.25
Step-by-step explanation:
Each person has equal probability of ⅓.
![\left[\begin{array}{cc}X&P(X)\\1&\frac{1}{3}\\2&\frac{1}{3}\\4&\frac{1}{3}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DX%26P%28X%29%5C%5C1%26%5Cfrac%7B1%7D%7B3%7D%5C%5C2%26%5Cfrac%7B1%7D%7B3%7D%5C%5C4%26%5Cfrac%7B1%7D%7B3%7D%5Cend%7Barray%7D%5Cright%5D)
The mean is the expected value:
μ = E(X) = ∑ X P(X)
μ = (1) (⅓) + (2) (⅓) + (4) (⅓)
μ = ⁷/₃
The standard deviation is:
σ² = ∑ (X−μ)² P(X)
σ² = (1 − ⁷/₃)² (⅓) + (2 − ⁷/₃)² (⅓) + (4 − ⁷/₃)² (⅓)
σ² = ¹⁴/₉
σ ≈ 1.25
