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mamaluj [8]
3 years ago
10

Simplify: -0.8b + 4.1c -(-3.2b) - 0.1c

Mathematics
2 answers:
Mars2501 [29]3 years ago
8 0

Answer:

2.4b+4c

Step-by-step explanation:

Simplify:

-0.8b+4.1c-(-3.2b)-0.1c

then;

-0.8b+4.1c+3.2b-0.1c

Like terms are those which have same variable to the same power.

Combine like terms;

2.4b+4c

Therefore, the simplified form of the given expression is, 2.4b+4c

umka21 [38]3 years ago
3 0
-0.8b+4.1c-(-3.2b)-0.1c
-(-3.2b)= 3.2b
4.1c-0.1c=4c
-0.8b+4c+3.2b
-0.8b+3.2b=2.4b

4c+2.4b
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19,552 views on day 20, 114,763 views cumulative total for 20 days.

Step-by-step explanation:

The decimal equivalent to "increased by 20% per day" would be (510)*(1.20)/day, if we started with 510 views as day 0.  Day 1 would be (510)*(1.20) = 612 views.

If we have n successively days, the equation would read (510)(1.20)^n.

This means that after, lets say, 3 days, the calculation would be (510)*(1.20)^3, or 881 views.

On day 20, the calculation is (510)*(1.2)^20, or 19,552 views.

But the question asks "How many total views did the video get over the course of the first 20 days . . ?"  This seems to be asking the sum of the first 20 days, which is easy if you use a spreadsheet.  But if you read the question to mean what are the total views on day 20, the answer would be 19,552 views.

If you read the question as asking for the total sum of views over 30 days, the answer is 114,763.  Quick:  what does that amount to at $0.05 per view.

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Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
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a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

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