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2 years ago

Which set or sets does the number 15 belong to

Mathematics

2 answers:

8_murik_8 [283]2 years ago

4 0

Rational,whole,integer,real

Andrews [41]2 years ago

4 0

The correct answers are:

Real; rational; integer; whole; natural

Explanation:

Real numbers cover most of the numbers we work with; they contain everything except complex (imaginary) numbers.

Rational numbers are any numbers that can be written as a fraction. 15 can be written as 15/1, which is a fraction, so it is a rational number.

Integers include positive and negative numbers, but no decimals or fractions. 15 is a positive number with no decimal, so it is an integer.

Whole numbers are positive numbers from 0 up, with no fractions or decimals. 15 is a positive number with no decimal, so it is a whole number.

Natural (counting) numbers are positive numbers from 1 up, with no fractions or decimals. 15 is a positive number with no decimal, so it is a natural number.

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2x - 20 = 32 20 - 3x = 8 6x - 8 = 16 -13 - 3x = -10

levacccp [35]

Answer:

Step-by-step explanation:

1st one is x=26

2nd one is x=4

3rd is x=4

4th is x=-1

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Mr. peter travels 130 miles in 2 hours. at this rate, how many miles will he travel in 30 minutes?

enyata [817]

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2 years ago

What is the probability that a 13-card bridge hand contains all cards from the same suit?

slega [8]

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2 years ago

there has to be at least one operating path. Once a component fails on a path, that path is no longer operating. The reliability

Eva8 [605]

Answer:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

Step-by-step explanation:

We can assume that the system work in series

If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

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2 years ago