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RUDIKE [14]
3 years ago
10

The cost of a layer cake is $17 plus $5 per layer. If Paul has only $30 to spend on a cake, what is the most number of layers he

can afford?
(Show how to solve it)
Mathematics
2 answers:
VikaD [51]3 years ago
5 0

Answer:

The most number of layers he can afford is 2

Step-by-step explanation:

We automatically have a fee of 17 so we know we have 13 dollars remaining. Using this information, we know that one layer is 5 dollars which means we can get 2 layers which is equal to 10 dollars which turns our total fee to $27. Hope this helps!

Gala2k [10]3 years ago
3 0
First off you have the starting cost (C).
C=$17

each layer is $5, so that becomes 5x.

the equation will be $17+5x=30

hope this helped !! :)
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At Billy's school, 80 students come to school by bicycle or by car. Together, the vehicles they arrive to school in have 270 whe
Anna35 [415]

Answer:

x = number of bicycles = 35

y = number of cars = 55

Step-by-step explanation:

Let

x = number of bicycles

y = number of cars

x + y = 80 (1)

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From (1)

x = 80 - y

Substitute x = 80 - y into (2)

2x + 4y = 270 (2)

2(80 - y) + 4y = 270

160 - 2y + 4y = 270

- 2y + 4y = 270 - 160

2y = 110

y = 110/2

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Substitute y = 55 into (1)

x + y = 80 (1)

x + 55 = 80

x = 80 - 55

x = 35

x = number of bicycles = 35

y = number of cars = 55

6 0
3 years ago
What the answer tonthis question 10x=-4(2x-9)​
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10x = 4(2x -9)

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10x - 8x = -36 - 8x

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If you need a better explanation just as in the comments.

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7 0
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There are three boxes: one with two golden coins, one with two silver coins, and one with one golden coin and one silver coin. A
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Answer:

Step-by-step explanation:

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{\displaystyle P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},}

where A and B are events and P(B) ≠ 0.

P(A) and P(B) are the probabilities of observing A and B without regard to each other.

P(A | B), a conditional probability, is the probability of observing event A given that B is true.

P(B | A) is the probability of observing event B given that A is true.

At this point, go through the attached file before you continue with part B.

Part B)

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note P(SS)=P(GG)=P(GS) = 1/3

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hence

P(Silver from SS) = 1/3

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