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Hatshy [7]
3 years ago
13

5(x - 3) = 2(x + 6) + 3x

Mathematics
1 answer:
Verdich [7]3 years ago
3 0
The first one is
5x-15=2x+12+3x
5x-2x-3x=12+15
0=27

The second one is
6.5x-7.75x=-12+14.5
-1.25x=-2.5
Divide both sides by - 1.25
X=. 02

The third one is
K-10=3
K=13
The second part of the third one is
-|k-10|=3
-k+10=3
-k=3+10
-k=13
Divide both sides by negative
K=-13

The fourth one is
4n+32=56
4n=56-32
4n=24
Divide both sides by 4
N=6
The second part to that one is
-4|n+8|=56
-4n-32=56
-4n=88
Divide both sides by - 4
N=22

The fifth one is
N+7-6=8
N=8-7+6
N=1+6
N=7
The second part to this one is
-|n+7|-6=8
-n-7-6=8
-n=8+7+6
-n=21
Divide both sides by negative
N=-21

The last one is
5x-7-3=9
5x=9+7+3
5x=19
Divide both sides by 5
X=19 over 5
The second part to this one is
-|5x-7|-3=9
-5x+7-3=9
-5x=9-7+3
-5x=5
Divide both sides by - 5
X=-1
I hope this helps you out.
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g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
3 years ago
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