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Marat540 [252]
3 years ago
14

The following observations on degree of polymerization for paper specimens are used to estimate the true average degree polymeri

zation The data has been given in an R vector called "data Copy and paste this code in R to create a boxplot and calculate the summary statistics Use the output and summary statistics to answer the following questions. data-c(418, 421, 421, 423, 425, 426, 432, 435, 437, 439, 445, 446, 449, 452, 454, 461, 464) boxplot(data, horizontal TRUE) mean(data) sd(data)
(a) Construct a boxplot of the data 420 430 440 450 460 420 425 430 435 440 445 450 455 0 420 430 440 450 460 o 420 430 440 450 460 Comment on any interesting features.
i) There is one outlier
ii) The data is strongly skewed
iii) There are no outliers
iv) The data appears to be centered near 438
v) The data appears to be centered near 428
vi) There is little or no skew
(b) Is it plausible that the given sample observations were selected from a symmetric distribution?
a) Yes
b) No
(c) Calculate a 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.) Tip: If you get one bound correct and the other incorrect this could mean that your method for calculation is incorrect. Recheck your calculations and your critical value Does the interval suggest that 437 is a plausible value for true average degree of polymerization? o Yes o No Does the interval suggest that 449 is a plausible value?
a) Yes
b) No
(d) Perform a hypothesis test to test the following hypotheses: ??: 1 # 449 The t test statistic is: The p-value is: (Round your answers to two decimal places.) o 0.001 < p-value < 0.002 o 0.002< p-value <0.01 o 0.01 p-value< 0.02 o 0.02 < p-value < 0.05 o 0.05 p-value 0.10 Does the hypothesis test suggest that 449 is a plausible value?
a) Yes
b) No
Mathematics
1 answer:
marshall27 [118]3 years ago
5 0

Answer:

The answer is in the explanation

Step-by-step explanation:

Mean value = \bar x = 438

(a) The data appeared to be centered towards 438.

There are little or no skew.

There is no outliers.

(b) yes

(c) standard deviation = s = 14.8955

95% confidence interval = +- t_{df}  * s\sqrt{n}

    438 +- 2.12 * 14.8955 /\sqrt{17}=

= (430.3415, 445.6585)

(d) Test statistic

t = (438 - 450)/{14.8955\sqrt17} }

t = -3.3216

p -value is correctly given in option b.

No, 450 is not a plausible valaue.

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Given the sequence -2,-8/3,-10/3,-4,-14/3 find the recursive formula
balu736 [363]

Step-by-step explanation:

-2, -8/3, -10/3, -4, -14/3

Write as multiples of 1/3.

-6/3, -8/3, -10/3, -12/3, -14/3

This is an arithmetic sequence where the first term is -6/3 and the common difference is -2/3.

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3 years ago
Prove divisibility 45^45·15^15 by 75^30
Anuta_ua [19.1K]

Answer:

3^{75}

Step-by-step explanation:

We are asked to divide our given fraction: \frac{45^{45}*15^{15}}{75^{30}}.

We will simplify our division problem using rules of exponents.

Using product rule of exponents (a*b)^n=a^n*b^n we can write:

45^{45}=(3*15)^{45}=3^{45}*15^{45}

75^{30}=(5*15)^{30}=5^{30}*15^{30}

Substituting these values in our division problem we will get,

\frac{3^{45}*15^{45}*15^{15}}{5^{30}*15^{30}}

Using power rule of exponents a^m*a^n=a^{m+n} we will get,

\frac{3^{45}*15^{45+15}}{5^{30}*15^{30}}

\frac{3^{45}*15^{60}}{5^{30}*15^{30}}

Using quotient rule of exponent \frac{a^m}{a^n}=a^{m-n} we will get,

\frac{3^{45}*15^{60-30}}{5^{30}}

\frac{3^{45}*15^{30}}{5^{30}}

Using product rule of exponents (a*b)^n=a^n*b^n we will get,

\frac{3^{45}*(3*5)^{30}}{5^{30}}

\frac{3^{45}*3^{30}*5^{30}}{5^{30}}

Upon canceling out 5^{30} we will get,

3^{45}*3^{30}

Using power rule of exponents a^m*a^n=a^{m+n} we will get,

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if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

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or using a unit vector for those above, then

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