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Nesterboy [21]
3 years ago
14

The total cost for a bucket of popcorn and 4 movie tickets is $56. The total cost for the same size bucket of popcorn and 6 movi

e tickets is $80. Which equation represents the relationship between y, the total cost of the popcorn and movie tickets, and x, the number of movie tickets that are purchased? y = 12x + 8 y = 12x + 24 y = 14x + 6 y = 14x + 28
Mathematics
1 answer:
Nat2105 [25]3 years ago
6 0
The answer is maybe the third one y=14x+6 
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Let X be the food expenditure of a family.
Let the population mean of an expenditure of a family be.
Let the population standard deviation of an expenditure of a family be.
The proportion families spend more than $30 but less than $490 per month on food is,
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3 years ago
Long division 5th grade I’m sorry if I’m asking a lot but I’m really new to this
Nuetrik [128]

Answer:

95.2

Step-by-step explanation:

4 0
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Read 2 more answers
Hey can you please help me posted picture of question
SpyIntel [72]
We are to find the Probability the someone buys a book that is paperback and fiction.

Let P(F) represents the event that the book is fiction and P(P) represents the event that the book is paperback. We are to find P(F∩P)

P(F∩P) = P(F) x P(P)

From the tree diagram we can see that:

P(F) = 0.45
P(P) = 0.65

Using the values, we get:

P(F∩P) = 0.45 x 0.65 = 0.2925

So, the Probability the someone buys a book that is paperback and fiction is 0.2925.

So, option B gives the correct answer
3 0
3 years ago
Enter the value of x 1.5 2.5 of a right triangle
vekshin1
If 1.5 and 2.5 are the height and base, then just use Pythagorean theorem,
7 0
3 years ago
What is the solution of
kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

\sqrt{2x+4}-\sqrt{x}=2

Isolating √(2x+4): Addind √x both sides of the equation:

\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}

Squaring both sides of the equation:

(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}

Simplifying on the left side, and applying on the right side the formula:

(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}

2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}

Squaring both sides of the equation:

(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

x^{2}-16x=16x-16x\\ x^{2}-16x=0

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2

x=0 is a solution


2) x=16

\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2

x=16 is a solution


Then the solutions are x=0 and x=16


5 0
3 years ago
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