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Nikolay [14]
2 years ago
15

Per

Mathematics
1 answer:
morpeh [17]2 years ago
6 0

Answer:

5 11/13km

Step-by-step explanation:

4 ( 1 6/13)

4× 19/13

= 5 11/13

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Convert y + 1 = 3/4 (x - 16) to standard form.
Archy [21]
3x/4-y=13 or choice B
7 0
3 years ago
A pound of turkey costs (3w 8) dollars and a pound of cheese costs (4w-3) dollars.mrs.young bought 2 pounds of turkey and three
zlopas [31]
x = 2(3w + 8) + 3(4w - 3) , where x is the total cost in terms of w.

x = (6w + 16) + (12w - 9)

x = 18w + 7
3 0
3 years ago
A mechanic has a set fee for working on someone's automobile and charges by the hour for making repairs. The total cost of her s
julia-pushkina [17]
I think it’s either b or c. But I’m leaning more towards c
7 0
3 years ago
Can someone PLEASE answer 5 and 6 for me?this assignment is due in a few minutes
Sergeu [11.5K]

Answer:

Yes they are equivalent because if you divide 18 by 6 it is 3 and if you do the same for 21 and 7 it also equals 3 which means they get paid 3 per hour

As for 6 they are not bec if we do the same from to 6 220/25=8.8 and 90/10=9 meaning they don’t have the same feet to inches

Step-by-step explanation:

I hope this helps and I’m not to late.

3 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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