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Alecsey [184]
2 years ago
12

Ayúdenme en esto porfavor​

Mathematics
1 answer:
soldier1979 [14.2K]2 years ago
7 0
<h3>¡Hola!</h3>

-----48

-----3/4

-----X

Sabiendo los datos

Planteamos ecuación:

==> En este caso debemos hallar el 3/4% de 48.

Usamos la regla de tres simple.

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf{ \boxed{\frac{3}{4} \times 48 }}

Resolvemos.

\sf{ \boxed{\frac{3}{4} \times 48 = \frac{ \frac{3}{4} }{100} \times 48 = \frac{3}{400} \times 48 = \frac{3}{25} \times 3 = \frac{9}{25} }}

Hallamos el decimal de 9/25

\:  \:  \:  \: \boxed{ \bold{\sf \rightarrow\frac{9}{25} = 9 \div 25 = 0.36 }}

Entonces podemos decir que:

En total a Susana le faltan 0.36km para llegar a casa.

#Spj2

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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.
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
Find all the real zeros of <br><br> f(x)=3x^3-9x^2+3x-9 <br><br><br> Can someone help me please
vaieri [72.5K]

Answer:

x=0,3

Step-by-step explanation:

start by dividing every term by 3

x^3-3x^2+x-3

group into 2 terms

(x^3-3x^2)+(x-3)

simplify as much as you can

x^2(x-3)+(x-3)

combine terms

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Answer:

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Step-by-step explanation:

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y=mx+c

-5 = -\frac{3}{2}(4)+c

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c=1

now we know the value of slope which is given -\frac{3}{2} and we found the value of which is 1 so we put these values into our original slope-intercept form equation

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quotient: division
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