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nevsk [136]
3 years ago
8

A 12 pack of cola costs $5.46. How much does one can of cola cost? In other words, find the unit rate of one can of cola.

Mathematics
2 answers:
Travka [436]3 years ago
5 0

Answer:

.45

Step-by-step explanation:

5.46 is a 12 pack

5.46/12 = .455

lapo4ka [179]3 years ago
3 0

Answer:

$0.46

Explanation:

12 cans = $5.46

1 can = .455

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Olegator [25]
Where is the question??? Bruh?
5 0
3 years ago
Which ordered pairs are in a proportional relationship with (0.2, 0.3)?
MAXImum [283]

Answer:

C and E

Step-by-step explanation:

We are given that

x_1=0.2,y_1=0.3

\frac{y}{x}=\frac{0.3}{0.2}=\frac{3}{2}

k=\frac{y}{x}=\frac{3}{2}

A.x_2=1.2,y_2=2.3

\frac{y_2}{x_2}=\frac{2.3}{1.2}=\frac{23}{12}\neq=\frac{3}{2}

Hence, it is not in proportional relationship with (0.2,0.3)

B.(2.7,4.3)

x_3=2.7,y_3=4.3

\frac{y_3}{x_3}=\frac{4.3}{2.7}=\frac{43}{27}\neq\frac{3}{2}

Hence, it is not in proportional relationship with (0.2,0.3).

C.(3.2,4.8)

x_4=3.2,y_4=4.8

\frac{y_4}{x_4}=\frac{4.8}{3.2}=\frac{3}{2}

Hence, the ordered pair (3.2,4.8) are in a proportional relationship with (0.2,0.3).

D.(3.5,5.3)

\frac{y_5}{x_5}=\frac{5.3}{3.5}=\frac{53}{35}\neq \frac{3}{2}

Hence, the ordered pair (3.5,5.3) are not in a proportional relationship with (0.2,0.3).

E.(5.2,7.8)

\frac{y_6}{x_6}=\frac{7.8}{5.2}=\frac{3}{2}

Hence, the ordered pair (5.2,7.8) are in a proportional relationship with (0.2,0.3).

4 0
3 years ago
26 points!!!
yanalaym [24]
It will be:
b=+ or - 5C
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bill and sam have a coupon for a pizza with 1 topping. The choices of toppings are: pepperoni, sausage, mushrooms, green peppers
cricket20 [7]
Well there are 6 toppings. For one person to select sausage, it is \frac{1}{6}. For two people, multiply them together and the probability is \frac{1}{36}
5 0
3 years ago
Xy + 3ey = 3e, find the value of y'' at the point where x = 0.
Katyanochek1 [597]

y might be 1 and I got that cuz 0y+3ey=3e and replace y with 1 and you get 0•1+3e•1=3e and to be honest it kinda looks like it makes sense

7 0
3 years ago
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