Answer:
a. z = 2.00
Step-by-step explanation:
Hello!
The study variable is "Points per game of a high school team"
The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)
The hypothesis is:
H₀: μ ≤ 99
H₁: μ > 99
α: 0.01
There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.
The statistic value under the null hypothesis is:
Z= X[bar] - μ = 101 - 99 = 2
σ/√n 6/√36
I don't have σ, but since this is an approximation I can use the value of S instead.
I hope it helps!
Hey
So my brother posted this on Yahoo
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm.
<span>What you do now is the following: </span>
<span>Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>
<span>So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8° </span>
<span>The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram. </span>
<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>≃ 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>
<span>b) </span>
<span>volume of water = cross-sectional area * length </span>
<span>≃ 13.5 * 30 cm³ </span>
<span>≃ 404 cm³</span>
Hoped it Helped
180 because you multiply the 12 packs by 10 to get 120. Then you subtract 120 from 300
