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N76 [4]
3 years ago
15

Jeremy bought a new pair of jeans. The jeans

Mathematics
2 answers:
Svet_ta [14]3 years ago
7 0

Answer:

28.80

Step-by-step explanation:

10% of $45 = 0.10 × 45 = $4.50

$45-$4.50=$40.50  

90% of $45 = 0.90 × 45 = $40.50

ivanzaharov [21]3 years ago
6 0
Pogba Varane Pogba est une bonne idée
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Write a second inequality with the same meaning -12<= a
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a\geq-12 or -48\leq4a

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The formula to find the volume of a any pyramid is:
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V = A(h/3)

Step-by-step explanation:

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3 years ago
Anyone know the answers to this ?!
torisob [31]

Answer:

Hey there!

3 x 5/4 = 3 x 5 ÷ 4 = 5 x 3 ÷ 4

5 x 4/3 = 5 x 4 ÷ 3 = 4 x 5 ÷ 3

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Hope this helps :)

5 0
3 years ago
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

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.

7 0
3 years ago
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