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adell [148]
3 years ago
14

10 points help due tomarrow

Mathematics
1 answer:
strojnjashka [21]3 years ago
4 0

Answer:

\boxed{0.2}

Step-by-step explanation:

Convert \dfrac{12}{60} to a decimal

Step 1. Reduce the fraction to its lowest terms.

Divide both numerator and denominator by their greatest common factor (12)

\dfrac{12}{60} =\dfrac{1}{5}

Step 2. Convert the denominator to a power of 10

Multiply both numerator and denominator by 2.

\dfrac{1}{5} \times \dfrac{2}{2} = \dfrac{2}{10}

Step 3. Divide the numerator by the denominator

Dividing by 10 moves the decimal point one place to the left.

\dfrac{2}{10} = \boxed{0.2}

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A a fraction close to but greater than 1 1/2
Bad White [126]

Answer:

Step-by-step explanation:

1 1/2 = 3/2

6 0
3 years ago
Es There are 30 students on the debate team and 20 students on the math team. Ten students are on both the math team and
irakobra [83]
There will be 40 students!!!
5 0
3 years ago
a computer and printer a totoal cost $1132 the cost of the computer is three times the cost of the ptinter
zavuch27 [327]
Use a system of equations

C+P=1132
3P=C

Substitute C in first equation as
3P+P=1132
Simplify
4P=1132
Solve
P=1132/4
P=283

NOW SOLVE FOR C SUBSTITUTING P VALUE IN FIRST EQUATION

C+283=1132
C=1132-283
C=849


Printer = 283$
Computer = 849$
3 0
2 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
PLEASE HELP!!!!! its Friday children uwu<br> this question isn't really math but yall smart !
Vesna [10]

Answer:

Mercury is smaller than Mars

8 0
2 years ago
Read 2 more answers
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