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Licemer1 [7]
3 years ago
13

If 3/5 of a chorus is singing. What fraction of the chorus is not singing?

Mathematics
2 answers:
Montano1993 [528]3 years ago
8 0
5/5 (one whole) - 3/5 = 2/5. 
Leokris [45]3 years ago
5 0
1- 3/5
= 5/5- 3/5
= 2/5

2/5 of the chorus is not singing~
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Which equation correctly applies the distributive property?
Elodia [21]

Answer:

−3⋅(6.48)=(−3⋅6)+(−3⋅0.4)+(−3⋅0.08)

Step-by-step explanation:

7 0
3 years ago
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What is the value of the digit 1 in the number 13?
MissTica

Answer:

Wouldn't it be the 10's place? I hope it's right.

Step-by-step explanation:

Ex. 136

1 = hundredths place

3 = tenths place

6 = ones place

8 0
3 years ago
Read 2 more answers
What is the value of r of the geometric series n=11.3(0.8)n-1
jeyben [28]

the value of r of the geometric series n=11.3(0.8)n-1

an=11.3(0.8)^{n-1}

General formula for nth term of any geometric series is   a_n=a_1(r)^{n-1}

Here 'r' is the common ratio

a_1 is the first term of the series

Now we compare the given formula with general formula

Compare  an=11.3(0.8)^{n-1}  with  a_n=a_1(r)^{n-1}

The value of r= 0.8


6 0
3 years ago
Three friends earned more than $200 washing cars. They paid their parents $28 for supplies and divided the rest of money equally
Annette [7]
Earned more than, so over $200. we don't know the EXACT amount, but whatever that amount is 28 dollars is being taken off. Then they split the money, so I write an inequality. (just think of it at a number line k)
       
     <----------O_____
__________

I couldn't really do better, sorry if I didn't help much.. :( but if I did pls leave a thanks :D
3 0
3 years ago
Determine n between 0 and 19 such that (2311)(3912) ≡ n mod 20.
sleet_krkn [62]

You can write 2311 and 3912 in the form 20q+r:

2311=115\cdot20+11

3912=125\cdot20+12

Then

2311\cdot3912=(115\cdot20+11)(125\cdot20+12)

2311\cdot3912=115\cdot125\cdot20^2+(11\cdot125+12\cdot115)\cdot20+11\cdot12

Taken modulo 20, the terms containing powers of 20 vanish and you're left with

2311\cdot3912\equiv11\cdot12\equiv132\pmod{20}

We further have

132=6\cdot20+12

so we end up with

2311\cdot3912\equiv12\pmod{20}

and so n=12.

###

If instead you're trying to find 2311^{3912}\pmod{20}, you can apply Euler's theorem. We can show that \mathrm{gcd}(2311,20)=1 using the Euclidean algorithm. Then since \varphi(20)=8, and 8 divides 3912, we have

2311^{3912}\equiv2311^{489\cdot8}\equiv(2311^{489})^8\equiv1\pmod{20}

To show 2311 and 20 are coprime:

2311 = 115*20 + 11

20 = 1*11 + 9

11 = 1*9 + 2

9 = 4*2 + 1   =>  gcd(2311, 20) = 1

3 0
4 years ago
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