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2 years ago

7

8.3.7 From the data on the pH of rain in Ingham County, Michigan: 5.47 5.37 5.38 4.63 5.37 3.74 3.71 4.96 4.64 5.11 5.65 5.39 4.

16 5.62 4.57 4.64 5.48 4.57 4.57 4.51 4.86 4.56 4.61 4.32 3.98 5.70 4.15 3.98 5.65 3.10 5.04 4.62 4.51 4.34 4.16 4.64 5.12 3.71 4.64 Find a two-sided 95% confidence interval for the standard deviation of pH.

1 answer:

Step2247 [10]2 years ago

3 0

Answer:

95% Confidence Interval: 4.7 ± 1.57

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In a gambling game, a woman is paid $3 if she draws a jack or a queen and $5 if she draws a king or an ace from an ordinary deck

cluponka [151]

There are 8 Jacks and Queens and 8 King and Aces

3*8+5*8=24+40=64 $

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2 years ago

a farmers packs 46 apples into trays for market.Each tray holds 6 apples.how many apples are in partially filled tray

denis23 [38]

$46 - 6 - 6 - 6 - 6 - 6 - 6-6=46-(7)(6)=46-42 = \boxed {4}$

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1 year ago

Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

###

Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":

$\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}$

which gives

$x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}$

as needed. Then with $0\le t\le2\pi$ , we compute the area via Green's theorem using the same setup as before:

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt$

$=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi$

3 0

2 years ago

The table shows the number of candies packed by Machine C. The equation shows the number of candies packed by Machine D. In both

Rashid [163]

Machine C makes 75 candies per minute.

Machine D makes 130 candies per minute.

Difference is 55

55 * 11 minutes = 605 candies.

4 0

1 year ago

Read 2 more answers

Given ΔMNO, find the measure of ∠LMN.

Novay_Z [31]

Answer:

104°

Step-by-step explanation:

If segments NO and NM are congruent, then angles NMO and NOM are congruent. So, their supplements, angles NML and NOP are congruent. That is ...

∠NML ≅ ∠NOP = 104°

∠NML = 104°

3 0

2 years ago

Read 2 more answers