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Whitepunk [10]
3 years ago
9

Wrapping paper is on sale at 3 rolls for 4.25. How many rolls can Danika buy with $20? (i've been searching all around the inter

net for the answer but i can't seem to find one similar to the answer my teacher gave me)
Mathematics
2 answers:
lana66690 [7]3 years ago
5 0
14 i’m pretty sure or 14.1
IRINA_888 [86]3 years ago
5 0

Answer:

Can buy 14 rolls

Step-by-step explanation:

Cost per roll:

  4.25 / 3 = 1.42 each roll

How many can buy for $20:

  1.42x = 20

  x = 14 rolls

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The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for t
nadezda [96]

Answer:

a. There is evidence that the population mean is above $300.

b. There is no evidence that the population mean is above $300.

c. There is no evidence that the population mean is above $300.

d. The director could ask for cheaper similar books.

Step-by-step explanation:

Let X be the random variable that represents the cost of textbooks. We have observed n = 25 values, \bar{x} = 315.4 and s = 43.20. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

H_{0}: \mu = 300 vs H_{1}: \mu > 300 (upper-tail alternative)

We will use the test statistic

T = \frac{\bar{X}-300}{S/\sqrt{25}} and the observed value is

t_{0} = \frac{315.4 - 300}{43.20/\sqrt{25}} = 1.7824.

If H_{0} is true, then T has a t distribution with n-1 = 24 degrees of freedom.

a. The rejection region is given by RR = {t | t > t_{0.9}} where t_{0.9} = 1.3178 is the 90th quantile of the t distribution with 24 df, so, RR = {t | t > 1.3178}. Because the observed value satisty 1.7824 > 1.3178, there is evidence that the population mean is above $300.

b. If s = 75, then the observed value is t_{0} = \frac{315.4 - 300}{75/\sqrt{25}} = 1.0267. The rejection region for a 0.05 level of significance is RR = {t | t > t_{0.95}} where t_{0.95} = 1.7108 is the 95th quantile of the t distribution with 24 df, so, RR = {t | t > 1.7108}. Because the observed value does not fall inside the rejection region, there is no evidence that the population mean is above $300.

c. If \bar{x} = 305.11 and s = 43.20, the observed value is t_{0} = \frac{305.11 - 300}{43.20/\sqrt{25}} =  0.5914. For RR = {t | t > 1.3178} we have that the observed value does not fall inside RR, therefore, there is no evidence that the population mean is above $300.

d. Because the director of admissions is concerned about the high cost of textbooks, and there is evidence that the population mean of costs is above $300, the director could ask for cheaper similar books.

8 0
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Step-by-step explanation:

∑ₙ₌₀°° n! xⁿ

Use ratio test:

L = lim(n→∞)│aₙ₊₁ / aₙ│

L = lim(n→∞)│[(n+1)! xⁿ⁺¹] / (n! xⁿ)│

L = lim(n→∞)│[(n+1)! / n!] (xⁿ⁺¹ / xⁿ)│

L = lim(n→∞)│(n+1) x│

L = ∞

The series is divergent for all values of x.

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Use ratio test:

L = lim(n→∞)│aₙ₊₁ / aₙ│

L = lim(n→∞)│[xⁿ⁺¹ / (n+1)!] / (xⁿ / n!)│

L = lim(n→∞)│[xⁿ⁺¹ / (n+1)!] (n! / xⁿ)│

L = lim(n→∞)│(xⁿ⁺¹ / xⁿ) [n! / (n+1)!]│

L = lim(n→∞)│x / (n+1)│

L = 0

The series is convergent for all values of x.

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