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3 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

Mathematics

2 answers:

yarga [219]3 years ago

8 0

Answer:

x = -2 ±i sqrt(6)

Step-by-step explanation:

x^2 +4x+10 = 0

We will complete the square

Subtract 10 from each side

x^2+4x +10-10 = -10

x^2+4x= -10

b = 4

Take b/2 and then square it

4/2 = 2 2^2 = 4

Add 4 to each side

x^2+4x+4= -10+4

The left side is rewritten as (x+b/2)^2

(x+2)^2 = -6

Take the square root of each side

sqrt((x+2)^2) = sqrt (-6)

x+2 = ±i sqrt(6)

Subtract 2 from each side

x+2-2 = -2 ±i sqrt(6)

x = -2 ±i sqrt(6)

Julli [10]3 years ago

4 0

Use the quadratic formula -b+/-√b^2 -4(ac) / 2a

-4 +/- √(4^2 - 4(1*10) / (2*1)

X = -2 +/- i√6

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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.

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