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Related questions with answers

Independent random samples were selected from populations 1 and 2. The sample sizes, means and variances are as follows:

Sample size population 1: 64 Sample size population 2: 64 Sample mean population 1: 2.9 Sample mean population 2: 5.1 Sample variance population 1: 0.83 Sample variance population 2: 1.67

a. Find a 90% confidence interval for the difference in population means. What does the phrase "90% confident" mean? b. Find a 99% confidence interval for the difference in the population means. Can you conclude that there is a difference in the two population means? Explain.

A cable lowers a $1200-\mathrm{kg}$ elevator so that the elevator’s speed increases from zero to $4.0 \mathrm{m} / \mathrm{s}$ in a vertical distance of $6.0 \mathrm{m}$ . What is the force that the cable exerts on the elevator while lowering it? Specify the system, its initial and final states, and any assumptions you made. Then change the system and solve the problem again. Do the answers match?

Perform the indicated operation. Write your answer in scientific notation. $\left( 8.9 \times 10 ^ { 7 } \right) \times \left( 9.8 \times 10 ^ { 10 } \right)$

Question

Perform the indicated operations, expressing answers in simplest form with rationalized denominators. $\frac{\sqrt{T^{4}-V^{4}}}{\sqrt{V^{-2}-T^{-2}}}$

Solution

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For the following expression:

\dfrac{\sqrt{T^4-V^4}}{\sqrt{V^{-2}-T^{-2}}}

To rationalize the denominator, multiply the numerator and the denominator by $\sqrt{V^{-2}-T^{-2}}$ as follows:

\ \ \ \ \ \ \ \ =\dfrac{\sqrt{T^4-V^4}}{\sqrt{V^{-2}-T^{-2}}} \times\dfrac{\sqrt{V^{-2}-T^{-2}}}{\sqrt{V^{-2}-T^{-2}}}

Apply this radical rule $\sqrt{a}\sqrt{a}=a$ as follows:

\ =\dfrac{\sqrt{T^4-V^4}\sqrt{V^{-2}-T^{-2}}}{V^{-2}-T^{-2}}

Simplify $V^{-2}-T^{-2}:\ \dfrac{1}{V^2}-\dfrac{1}{T^2}$ using this rule $a^{-n}=\dfrac{1}{a^n}$ as follows:

=\dfrac{\sqrt{T^4-V^4}\sqrt{ \frac{1}{V^2}-\frac{1}{T^2}}}{\frac{1}{V^2}-\frac{1}{T^2}}

Combine the fractions $\dfrac{1}{V^2}-\dfrac{1}{T^2}=\dfrac{T2-V^2}{V^2T^2}$ as follows:

\ \ =\dfrac{\sqrt{T^4-V^4}\sqrt{\dfrac{T^2-V^2}{T^2V^2}}}{\dfrac{T2-V^2}{V^2T^2}}

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