It takes 5.22 hours to install carpet in a room. Step-by-step explanation: To start, let's explore the relationship among yards, feet, and inches. 1. 1 yard = 3 feet. 2. 1 foot = 12 inches. 3. Therefore, 1 yard = 36 inches. Now, let's convert all measurements to inches. ∵ 1 yard = 36 inches ∴ 1 yard² = (36)² inches². - Fine Floors can outfit 15 square yards of carpeting in 4 hours and 30 minutes. ∵ 15 yard² equals 15 × (36)² inches², so 15 yards² = 19440 inches². ∵ 1 hour equals 60 minutes, thus 30 minutes = 0.5 hours. Consequently, 4 hours and 30 minutes = 4.5 hours, implying Fine Floors covers 19440 inches² in 4.5 hours. The dimensions of the room are 11 feet 9 inches by 13 feet 4 inches. ∵ 1 foot = 12 inches, therefore, 11 feet 9 inches = 11 × 12 + 9 = 141 inches, and 13 feet 4 inches = 13 × 12 + 4 = 160 inches. Hence, the room’s area is 141 × 160 = 22560 inches². Using the ratio method, we have area of carpeting (in²): time of carpeting (hr) hence, 19440: 4.5 = 22560: h. By cross-multiplying, 19440h = 4.5 × 22560 yields 19440h = 101520. Dividing both sides by 19440 results in h = 5.22 hrs.
To determine the values of b that fulfill 3(2b+3)^2 = 36
we start with
3(2b+3)^2 = 36
Divide both sides by 3
(2b+3)^2 = 12
Next, take the square root of both sides
(2b+3)} = (+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3
b1=\frac{\sqrt{12}}{2} -\frac{3}{2}
b1=\sqrt{3} -\frac{3}{2}
b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
Thus,
the solutions for b are
b1=\sqrt{3} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
The attached graph illustrates the region. The centroid's coordinates are (5/3, 1). The centroid's coordinates are determined by averaging the coordinates of the area; Oₓ = (Aₓ+Bₓ+Cₓ)/3 = (0+1+4)/3 = 5/3 and O(y) = (A(y) + B(y) + C(y)) = (0+3+0)/3=3/3=1.
Answer:
At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.
Step-by-step explanation:
A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.
The null and alternative hypotheses are:
The significance level is set at 0.10.
The sample mean recorded is 17, and the sample standard deviation is 5.37.
The degrees of freedom are calculated as df=(20-1)=19.
The t-statistic is:
The two-tailed P-value corresponding to t=1.67 is P=0.11132.
<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.
At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.
