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alukav5142 [94]
3 years ago
10

1. 35, 56, 34, 44, 52, 12, 34, 45 Mean: Median: Mode:

Mathematics
2 answers:
Naily [24]3 years ago
7 0

Mean: 39

Median: 39.5

Mode: 34

kherson [118]3 years ago
5 0

Answer:

Mean x¯¯¯ 39

Median x˜ 39.5

Mode 34

Step-by-step explanation:

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point slope form is y-y1=m(x-x1)

from (6,1) y1 = 1 and x1 = 6

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Step-by-step explanation:

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Plzzzzzz someone help I keep on getting a decimal answer that is 100% wrong. Please do number 14
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Which points are on the graph of g(x)=(1/5)x<br><br> Select EACH CORRECT ANSWER
sdas [7]

If a point is on the graph of the exponential equation then it should be true when you solve.


g(x) = (1/5)^x

y = (1/5)^x


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6 0
3 years ago
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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
3 years ago
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