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Questions in mathematics

[Answered] Simplify: [tex]$\frac{x}{x^2-9}+\frac{3}{9-x^2}$[/tex]

[Answered] Given that $u=3-2 x^2$, which integral is equivalent to $\int_{-1}^0 x^3 \sqrt{3-2 x^2} d x $? a $\quad \int_{-1}^0(3-u) \sqrt{u} d u$ b $\frac{1}{8} \int_1^3(3-u) \sqrt{u} d u$ c $\frac{1}{4} \int_{-1}^0(3-u) \sqrt{u} d u$ d $\frac{1}{2} \int_1^3(3-u) \sqrt{u} d u$ e $-\frac{1}{8} \int_1^3(3-u) \sqrt{u} d u$

[Answered] Given two dependent random samples with the following results: | Population 1 | 30 | 47 | 19 | 49 | 42 | 31 | 24 | |---|---|---|---|---|---|---|---| | Population 2 | 45 | 37 | 29 | 44 | 46 | 46 | 34 | Use this data to find the $90 \%$ confidence interval for the true difference between the population means. Assume that both populations are normally distributed. Step 1 of 4: Find the point estimate for the population mean of the paired differences. Let $x_1$ be the value from Population 1 and $x_2$ be the value from Population 2. Use the formula $d=x_2-x_1$ to calculate the paired differences. Round your answer to one decimal place.

[Answered] Which equation represents a line that passes through ($4, \frac{1}{3}$) and has a slope of $\frac{3}{4}$? A. $y-\frac{3}{4}=\frac{1}{3}(x-4)$ B. $y-\frac{2}{3}=\frac{3}{4}(x-4)$ C. $y-\frac{1}{3}=4\left(x-\frac{3}{4}\right)$ D. $y-4=\frac{3}{4}\left(x-\frac{1}{3}\right)$

[Answered] Consider the function [tex]f(x)=12 x^5+30 x^4-300 x^3+3[/tex]. [tex]f(x)[/tex] has inflection points at (reading from left to right) [tex]x=D, E[/tex], and [tex]F[/tex] where [tex]D[/tex] is $\square$ and [tex]E[/tex] is $\square$ and [tex]F[/tex] is $\square$

[Answered] Use the guidelines for graphing to make a complete graph of [tex]$f(x)=3 x-5 \sqrt{x}$[/tex] Start by identifying the domain or interval of interest. The domain of [tex]$f$[/tex] is ____. (Type your answer in interval notation.)

[Answered] Expand $\left(x-\frac{1}{3} y+\frac{1}{5} z\right)^2$

[Answered] Given two dependent random samples with the following results: | Population 1 | 30 | 47 | 19 | 49 | 42 | 31 | 24 | |---|---|---|---|---|---|---|---| | Population 2 | 45 | 37 | 29 | 44 | 46 | 46 | 34 | Use this data to find the $90 \%$ confidence interval for the true difference between the population means. Assume that both populations are normally distributed. Step 2 of 4 : Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.

[Answered] Solve for $x$: $-2.863 .98 - 0.8y = -2.86$

[Answered] Evaluate the following. $3 \times 5+9+25 \div 5$