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vekshin1
1 year ago
11

How can data displays be used to compare two sets of data? Select all that apply. They quickly illustrate measures of center. Th

ey quickly illustrate measures of center. They identify all the data points in each data set. They identify all the data points in each data set. They show trends in data that can be compared. They show trends in data that can be compared. They display measures of variability for each data set. They display measures of variability for each data set. They easily show the mean of each data set.
Mathematics
1 answer:
Virty [35]1 year ago
6 0

Answer:

Step-by-step explanation:

A and C Step-by-step explanation:I would go for A and C options which best describes the use of data displays for the comparison of two data sets.Data Display is basically displaying of useful data which has been extracted from bundles of raw data or raw data points. That useful data can be used to compare two different datasets as well. So, in option A, It says that it quickly illustrate measures of centre. True, because it presents you quick display so that everyone seeing the data in form of charts or tables easily catch the information to be conveyed. And in option C, It says, they show trends in data that can be compared. Yes again true. Data displays show you trends of different things in one clear picture and it can be compared with other datasets through the use of data displays. Still stuck? Get 1-on-1 help from an expert tutor now.

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Step-by-step explanation:

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2 years ago
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How do I solve these problems? ln(x) = 5.6 + ln(7.5) and log(x) = 5.6 - log(7.5)
SOVA2 [1]

Use the rules of logarithms and the rules of exponents.

... ln(ab) = ln(a) + ln(b)

... e^ln(a) = a

... (a^b)·(a^c) = a^(b+c)

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1) Use the second rule and take the antilog.

... e^ln(x) = x = e^(5.6 + ln(7.5))

... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents

... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms

... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)

2) Similar to the previous problem, except base-10 logs are involved.

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8 0
3 years ago
Solve for x ....... thanks !
Len [333]
If we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b

16=2⁴
32=2⁵
rememeber that (x^m)^n=x^{mn}

16^{3x+2}=32^{-2x-7}
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2^{4(3x+2)}=2^{5(-2x-7)}
2=2 so we conclude that 4(3x+2)=5(-2x-7)

4(3x+2)=5(-2x-7)
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Answer:

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