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prisoha [69]
3 years ago
13

Uring a field trip, Liu notices that an old marble statue looks different from a new marble statue. The surface of the old statu

e is rough while the new statue is smooth and shiny. There is a high level of pollution in the area the statues are located. What is most likely the cause of the difference in the way the statues appear?
Mathematics
1 answer:
Alja [10]3 years ago
5 0

Answer:

Liu notice that the old marble statue looks different from the new one also the old marble statue is rough while the new marble statue is smooth and shiny. Chemically marble is mixture of majorly  Calcium oxide, Silica ,Alumina and other salts.

Since the environment is highly polluted so it enables chemical reactions on the surface of the statues and that is what causing the degradation of the statues to loose their shine and smoothness

Hence,  the new statues looked smooth and shiny while the old statues looked rough and dull.  

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4\sqrt{15}

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39xy 6xy2 3x2y3 What is the greatest common factor (GCF) of the monomials shown above?
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2x^2  y^3

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2x^2 / x^2 - y^2 - x^2 / xy - y ^ 2​
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2 - 2y^2 - x/y

Step-by-step explanation:

2x^2/x^2 = 2

and

x^2/xy = x/y

therefore

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3 years ago
The Pew Internet and American Life Project reported Wednesday, April 18th that two- thirds (67%) of young adults with profiles o
dlinn [17]

Answer:

At 5% significance level, larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.

Step-by-step explanation:

let p be the proportion of military personnel students who restrict access to their profiles. Then null and alternative hypotheses are:

H_{0}: p=0.67 (67%)

H_{a}: p>0.67

We need to calculate z-statistic of sample proportion:

z=\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } } where

  • p(s) is the sample proportion of military students who restrict access to their profiles ( \frac{78}{100} =0.78)
  • p is the proportion assumed under null hypothesis. (0.67)
  • N is the sample size (100)

Then z=\frac{0.78-0.67}{\sqrt{\frac{0.67*0.33}{100} } } ≈ 2.34

The corresponding p-value is 0.0096. Since 0.0096<0.05 (significance level) we can reject the null hypothesis and conclude at 5% significance level that larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.

3 0
3 years ago
What is the simplified form of the expression? 7^4
nalin [4]
D)2401
7^4=7^2×7^2=49×49=2401
4 0
3 years ago
Read 2 more answers
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