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Katyanochek1 [597]

2 years ago

8

choose all statements about percentages that are true a. 15% of 30 is 12 b. 25% of 80 is 20 c. 30% of 50 is 25 d. 30% of 60 is 1

8 e. 40% of 20 is 8

1 answer:

Mashcka [7]2 years ago

7 0

Answer:a b c

Step-by-step explanation:

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A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ag

Katyanochek1 [597]

Answer:

At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll

Step-by-step explanation:

Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.

Let five years ago be group I X and as of now be group II Y

$H_0: p_x =p_y\\H_a: p_x \neq p_y$

(Two tailed test at 5% level of significance)

Group I Group II combined p

n 270 300 570

favor 120 140 260

p 0.4444 0.4667 0.4561

Std error for differene = $\sqrt{\frac{0.4561(1-0.4561)}{570} } \\=0.0209$

p difference = -0.0223

Z statistic = p diff/std error = -1.066

p value =0.2864

Since p value >0.05, we accept null hypothesis.

At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll

7 0

3 years ago

I think of a number multiply by three add four I get 22. Using the statement above form an equation use the letter x for the unk

bogdanovich [222]

Answer:

x=6

Step-by-step explanation:

(x*3)+4=22

Subtracting 4 from both sides:

(x*3)=18

Dividing both sides by 3:

x=6

Hope this helps!

4 0

2 years ago

Pls help me guys!<br> Thanks =P

Tema [17]

(4*7/2)+(15*7)+(7*6/2) = 140

6 0

2 years ago

The average earning of a shop for first three weeks of a month are 2750 per week. If the earnings of next 2 weeks are 4750 rupee

lisabon 2012 [21]

Answer:

D. 13000 is the correct option out of the choices!

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

2 years ago