Which equation is not equvilant It is c
Repeating decimal representation will be 3/13, 7/24. Terminating decimal representation will be 3/16, 4/125.
<h3>What is decimal?</h3>
The accepted method for representing both integer and non-integer numbers is the decimal numeral system. It is the expansion of the Hindu-Arabic numeral system to non-integer values. Decimal notation is the term used to describe the method of representing numbers in the decimal system. The base-10 number system, arguably the most widely used number system, is referred to as decimal. The decimal number system consists of ten single-digit numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. The number following nine is 10.
Here,
Repeating decimal representation,
=3/13, 7/24
Terminating decimal representation,
=3/16, 4/125
3/13 and 7/24 will be repeated in decimal form. The final two decimal places are 3/16 and 4/125.
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Answer:
46
Step-by-step explanation:
Let
represent a number with
in the ten's position and
is the one's position.
This means
actually has value of
.
We are given the sum of those digits of
is 10; this means
.
It says if 18 is added to the number
, then the result is
.
So
has value
and
has value
.
We are given then:


Subtract
on both sides:

Simplify:

Subtract
on both sides:


Divide both sides by 9:

Rearrange by commutative property:

So the system of equations we want to solve is:


-------------------------Add equations together (this will eliminate the variable
and allow you to go ahead and solve for
:


Divide both sides by 2:

Simplify:

If
and
, then
.
since 4+6=10.
So the original number is (46).
18 more than 46 is 18+46=(64) which is what we wanted.
We also have the sum of 4 and 6 is 10 as well.
Answer:
A, C are true . B is not true.
Step-by-step explanation:
Mean of a discrete random variable can be interpreted as the average outcome if the experiment is repeated many times. Expected value or average of the distribution is analogous to mean of the distribution.
The mean can be found using summation from nothing to nothing x times Upper P (x) , i.e ∑x•P(x).
Example : If two outcomes 100 & 50 occur with probabilities 0.5 each. Expected value (Average) (Mean) : ∑x•P(x) = (0.5)(100) + (0.5)(50) = 50 + 25 = 75
The mean may not be a possible value of the random variable.
Example : Mean of possible no.s on a die = ( 1 + 2 + 3 + 4 + 5 + 6 ) / 6 = 21/6 = 3.5, which is not a possible value of the random variable 'no. on a die'