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Arturiano [62]
1 year ago
14

8) The height of a building h varies directly as the length and inversely as the width, if h = 25m when l = 10m and w = 5m, find

the width when h = 50m and I remains the same
Mathematics
1 answer:
shtirl [24]1 year ago
8 0

Let h,l, and w represent the height, length, and width of the building respectively

Then

\begin{gathered} h\text{ = }\frac{kl}{w} \\ \text{where k is the constant of proportionality} \end{gathered}\begin{gathered} \Rightarrow k=\frac{hw}{l} \\ \Rightarrow k=\frac{25\times5}{10}=12.5 \end{gathered}\Rightarrow h=\frac{12.5l}{w}

The new value of h = 50m, but l remains the same.

That is l = 10m

Therefore

\begin{gathered} \frac{50}{1}=\frac{12.5\times10}{w} \\ \Rightarrow\frac{50}{1}=\frac{125}{w} \\ \text{Cross multiplying, we have} \\ 50w\text{ = 125} \\ \Rightarrow\text{ w=}\frac{125}{50}=2.5 \end{gathered}

Hence, the width is 2.5m

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So, we are given two sites;

Site one: 91.28 , 92.83, 89.35, 91.90 82.85, 94.83, 89.83, 89.00, 84.62, 86.96, 88.32, 91.17, 83.86, 89.74, 92.24, 92.59, 84.21, 89.36, 90.96, 92.85, 89.39, 89.82, 89.91, 92.16, 88.67.

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STEP ONE: DETERMINE THE DEGREE OF FREEDOM AND THE CRITICAL VALUE.

The degree of freedom =( mean of site 1 ) + (mean of site 2) - 2 = (25 + 25) − 2 = 48.

Since the degree of freedom is 48, Recall that the value for alpha given = 0.02. Then, the critical value = 2.407.

STEP TWO: DETERMINE THE STANDARD DEVIATION AND THE STANDARD ERROR FOR BOTH.

Therefore, the standard deviation = √[( 25 - 1) × 3.005^2 + (25 -1) × 3.271^2) ÷ (25 +25 - 2) ] = 3.14.

Also, the standard error = standard deviation × ( √(1/ mean 1 + 1/mean 2) ..

Standard error = 3.14 × [ ✓( 1/25 + 1/25)] = 0.89.

STEP THREE: DETERMINE THE CONFIDENCE INTERVAL.

confidence interval = (89.548 - 89.033 - 2.41 × 0.89, 89.548 - 89.033 + 2.41 × 0.89) = (- 1.62, 2.65).

STEP FOUR: THE HYPOTHESES.

=> THE TWO TAILED TEST.

Jo:μ1​ = μ2​.

Ja: μ1​ ≠ μ2​.

Thus, the rejected region is {t : ∣ t ∣ > 2.7}

As critical value = 2.7 for the degree of freedom which is 48 and α=0.01.

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(89.548 - 89.003)÷ ( standard deviation × √ ( 1/25 + 1/25). = 0.58.

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Also, from the p-value table which gives p=0.565. The p- value is greater than or equal to the alpha value(0.01). therefore, the null hypothesis is not rejected.

We are not going to reject the null hypothesis because the information gathered is not enough.

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