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igor_vitrenko [27]
3 years ago
5

Idk what I'm doing at all and I'm too tired to focus plz help​

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
7 0

Independent variables don't rely on anything. This will be $20. You pay $20 regardless.

Dependent variables will affect the outcome. This will be the number of friends. If f (the number of friends) ≥ 5, than you pay $3 per person.

c = cost

if f < 5

c = 20

if f ≥ 5,

c = 20 + 3f

Table

f c($)

5 35

6 38

7 41

8 44

9 47

10 50

11 53

12 56

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Answer:

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6 0
4 years ago
The areas of two similar figures are 20 cm^2 and 45 cm^2. What is the scale factor of their corresponding sides?
ra1l [238]

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3 years ago
Please help with this question<br> 2(2w-6)=6(w-6)+14<br> w=?
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8 0
3 years ago
A bag contains 4 red balls, 2 green balls, 3 yellow balls, and 5 blue balls. Find each probability for randomly removing balls w
AfilCa [17]

Answer:

\frac{15}{4802}, \frac{15}{9604}, \frac{9}{2401}, \frac{9}{4802}

Step-by-step explanation:

The bag has a total of (4+2+3+5) = 14 balls. Set up the proportions:

Red: \frac{4}{14}

green: \frac{2}{14}

yellow: \frac{3}{14}

blue: \frac{5}{14}

Now solve!

Removing 1 yellow, 1 red, 1 green, and 1 blue = \frac{3}{14} \cdot \frac{4}{14}  \cdot \frac{2}{14}  \cdot \frac{5}{14} =\frac{15}{4802}

Removing 1 blue, 1 green, 1 green, and 1 yellow = \frac{5}{14} \cdot \frac{2}{14}  \cdot \frac{2}{14}  \cdot \frac{3}{14} =\frac{15}{9604}

Removing 1 red, 1 red, 1 yellow, and 1 yellow = \frac{4}{14} \cdot \frac{4}{14}  \cdot \frac{3}{14}  \cdot \frac{3}{14} =\frac{9}{2401}

Removing 1 green, 1 yellow, 1 yellow, and 1 red = \frac{2}{14} \cdot \frac{3}{14}  \cdot \frac{3}{14}  \cdot \frac{4}{14} =\frac{9}{4802}

4 0
2 years ago
I need help :3 !!!!!!
Vikki [24]

Answer:

b = 43

Since 43 is congruent to b

4 0
3 years ago
Read 2 more answers
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