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3 years ago

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According to a recent study, the mean cost of a 3 bedroom home costs less in Ohio than in Alabama. You think this information is

incorrect. You randomly select 35 homes in each state. In Ohio, the mean cost is $137,400. In Alabama, the mean cost is $135,900. From past studies, the population standard deviation for 3 bedroom homes in Ohio is $3200 and in Alabama is $2800. Is there enough evidence to support your claim at α = 0.05?

1 answer:

Kobotan [32]3 years ago

3 0

Answer:Yes it is

Step-by-step explanation:

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How do I find The first figure is dilated to form the second figure.

galben [10]

Answer:

The marked choice is correct

Step-by-step explanation:

The scale factor is the ratio of the dimensions of the second figure to those of the first:

1.45/5.8 = 0.25

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4 years ago

Use the substitution x = 2 − cos θ to evaluate the integral ∫ 2 3/2 ( x − 1 3 − x )1 2 dx. Show that, for a < b, ∫ q p ( x −

MrRissso [65]

If the integral as written in my comment is accurate, then we have

$I=\displaystyle\int_{3/2}^2\sqrt{(x-1)(3-x)}\,\mathrm dx$

Expand the polynomial, then complete the square within the square root:

$(x-1)(3-x)=-x^2+4x-3=1-(x-2)^2$

$I=\displaystyle\int_{3/2}^2\sqrt{1-(x-2)^2}\,\mathrm dx$

Let $x=2-\cos\theta$ and $\mathrm dx=\sin\theta\,\mathrm d\theta$ :

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-(2-\cos\theta-2)^2}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-\cos^2\theta}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{\sin^2\theta}\sin\theta\,\mathrm d\theta$

Recall that $\sqrt{x^2}=|x|$ for all $x$ , but for all $\theta$ in the integration interval we have $\sin\theta>0$ . So $\sqrt{\sin^2\theta}=\sin\theta$ :

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sin^2\theta\,\mathrm d\theta$

Recall the double angle identity,

$\sin^2\theta=\dfrac{1-\cos(2\theta)}2$

$I=\displaystyle\frac12\int_{\pi/3}^{\pi/2}(1-\cos(2\theta))\,\mathrm d\theta$

$I=\dfrac\theta2-\dfrac{\sin(2\theta)}4\bigg|_{\pi/3}^{\pi/2}$

$I=\dfrac\pi4-\left(\dfrac\pi6-\dfrac{\sqrt3}8\right)=\boxed{\dfrac\pi{12}+\dfrac{\sqrt3}8}$

You can determine the more general result in the same way.

$I=\displaystyle\int_p^q\sqrt{(x-a)(b-x)}\,\mathrm dx$

Complete the square to get

$(x-a)(b-x)=-(x-a)(x-b)=-x^2+(a+b)x-ab=\dfrac{(a+b)^2}4-ab-\left(x-\dfrac{a+b}2\right)^2$

and let $c=\frac{(a+b)^2}4-ab$ for brevity. Note that

$c=\dfrac{(a+b)^2}4-ab=\dfrac{a^2-2ab+b^2}4=\dfrac{(a-b)^2}4$

$I=\displaystyle\int_p^q\sqrt{c-\left(x-\dfrac{a+b}2\right)^2}\,\mathrm dx$

Make the following substitution,

$x=\dfrac{a+b}2-\sqrt c\,\cos\theta$

$\mathrm dx=\sqrt c\,\sin\theta\,\mathrm d\theta$

and the integral reduces like before to

$I=\displaystyle\int_P^Q\sqrt{c-c\cos^2\theta}\,\sin\theta\,\mathrm d\theta$

where

$p=\dfrac{a+b}2-\sqrt c\,\cos P\implies P=\cos^{-1}\dfrac{\frac{a+b}2-p}{\sqrt c}$

$q=\dfrac{a+b}2-\sqrt c\,\cos Q\implies Q=\cos^{-1}\dfrac{\frac{a+b}2-q}{\sqrt c}$

$I=\displaystyle\frac{\sqrt c}2\int_P^Q(1-\cos(2\theta))\,\mathrm d\theta$

(Depending on the interval [<em>p</em>, <em>q</em>] and thus [<em>P</em>, <em>Q</em>], the square root of cosine squared may not always reduce to sine.)

Resolving the integral and replacing <em>c</em>, with

$c=\dfrac{(a-b)^2}4\implies\sqrt c=\dfrac{|a-b|}2=\dfrac{b-a}2$

because $a$ , gives

$I=\dfrac{b-a}2(\cos(2P)-\cos(2Q)-(P-Q))$

Without knowing <em>p</em> and <em>q</em> explicitly, there's not much more to say.

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3 years ago

Add the two expressions:

Korvikt [17]

(-6.3x + 4) + (1.5x - 6)

-6.3x + 1.5x = -4.8x
4 + (-6) = -2

-4.8x - 2 is your answer

hope this helps

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3 years ago

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evablogger [386]

Answer:

5.1

Step-by-step explanation:

Add 4.3 to 0.8

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4 years ago

Which value is equivalent to the expression 32 × 33?

PilotLPTM [1.2K]

Answer:

The correct answer is D) 243

Step-by-step explanation:

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3 years ago