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Elis [28]
3 years ago
5

Is .3333333333333 a rational number??

Mathematics
2 answers:
natali 33 [55]3 years ago
8 0

Answer: No

Step-by-step explanation: Because it is.

loris [4]3 years ago
7 0

Answer:

No .3333333333333 is a rational number

Explanation:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number.

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Six times the amount of women that were at the banquet last year will be in attendance this year. The same amount of men will be
Rina8888 [55]

Answer:

m+6n

Step-by-step explanation:

Let number of men who attended last year=m

Let number of women who attended last year=n

This year, we expect Six times the amount of women attendance last year =6 X n = 6n

The same number of men, m will be expected this year

Therefore, Number of Expected Attendees

= Number of Men + (Six Times the Number of Women)

=m+6n

We can expect (m+6n) people to be in attendance this year.

5 0
3 years ago
How many terms are in this expression? d + 8c + 7 ​
Fantom [35]

Answer:

2

Step-by-step explanation:

8 0
3 years ago
Which inequality has a solution that is the set of all real numbers?
kati45 [8]

Answer:

12<14

x<12

14>12

14>x

Step-by-step explanation:

I dont know if it is correct but I think it is that way try

6 0
3 years ago
I have a past due balance of $87.50 and my new charges are $75.00. I was also charged a late fee of 15% of my past due balance.
Kitty [74]
Past due balance = $87.50
Late fee = 15% of $87.50 = 0.15*87.50 = $13.125
New charges = $75.00

New total = 87.50 + 13.125 + 75 = $175.625

Answer: The new total is $175.63
8 0
3 years ago
Read 2 more answers
A bin contains 25 light bulbs, 5 of which are in good condition and will function for at least 30 days, 10 of which are partiall
Ira Lisetskai [31]

Answer:

The probability that it will still be working after one week is \frac{1}{5}

Step-by-step explanation:

Given :

Total number of bulbs = 25

Number of bulbs which are good condition and will function for at least 30 days = 5

Number of bulbs which are partially defective and will fail in their second day of use = 10

Number of bulbs which are totally defective and will not light up = 10

To find : What is the probability that it will still be working after one week?

Solution :

First condition is a randomly chosen bulb initially lights,

i.e. Either it is in good condition and partially defective.

Second condition is it will still be working after one week,

i.e. Bulbs which are good condition and will function for at least 30 days

So, favorable outcome is 5

The probability that it will still be working after one week is given by,

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

\text{Probability}=\frac{5}{25}

\text{Probability}=\frac{1}{5}

5 0
3 years ago
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