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Pani-rosa [81]
3 years ago
6

Help with this word problem???? :)

Mathematics
1 answer:
ycow [4]3 years ago
4 0

Answer:

I'm pretty sure it's 144

Step-by-step explanation:

1.           160 divided by 5 is 32 so 1 quart = 32 pounds

2.          32 times 4.5 = 144

P.S Mark as brainliest rate 5 stars and hit that thanks button!

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Two baseball teams, the Jayhawks and the Falcons, each recorded the number of home runs hit in 1 season by every player on the t
mihalych1998 [28]
The right answer for the question that is being asked and shown above is that: "B. The overlap is high because the difference in the means is small compared to either MAD" The <span>statement that best describes the overlap in the distribution of the two data sets is that </span><span>B. The overlap is high because the difference in the means is small compared to either MAD</span>
5 0
3 years ago
How would I finish solving this?
Zarrin [17]
Hello!

The original equation is:

200 = (5w ÷ 2)8

This problem can be written as:

200 = (5w/2)8

And you can then reduce the numbers with 2:

200 = 5x * 4
200 = 20x
10 = x

Your correct answer is 10.
3 0
3 years ago
Read 2 more answers
Solve the equation for n and show your work. -3n + 48 = 0
MrRissso [65]
-3n+48=0
-3n=-48
-3n/-3 & -48/-3
n=16
7 0
2 years ago
I got A, B, D, E. Can I have help with confirmation?
Margarita [4]

Answer:

Thats correct i think°

7 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
3 years ago
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