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Dafna1 [17]
3 years ago
11

How many solutions are there to the equation below?

Mathematics
2 answers:
Anton [14]3 years ago
8 0

Answer:

B. Infinitely Many

Step-by-step explanation:

Step 1: Distribute

5x + 48 + 7x = 12x + 48

Step 2: Combine like terms

12x + 148 = 12x + 48

If the 2 lines are the same, which they are, they are the same lines. That means there are an infinite amount of solutions. If they had the same slope but different y-intercepts, there would be no solutions because the lines are parallel. Any other pair of equations would have 1 solution.

Fudgin [204]3 years ago
6 0

Answer:

B. Infinitely Many

Step-by-step explanation:

When you simplify the term on the left of the equation, it is exactly the same as the term on the right. Therefore, you can substitute virtually anything in for variable "x" and it would still be correct.

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24% of U.S. adults say they are more likely to make purchases during a sales tax holiday. you randomly select 10 adults. find th
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Answer:

a) p(X= 2) = 0.261

b) P(x>2) = 0.566

c) P(2<x<5)  = 0.334

Step-by-step explanation:

Given 24% of U.S. adults say they are more likely to make purchases during a sales tax holiday

Probability 0f U.S. adults say they are more likely to make purchases during a sales tax holiday (p) = 0.24

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a)

The probability of getting exactly '2'

The probability p(X= 2) = e^{-\alpha } \frac{\alpha^r }{r! }

p(X= 2) = e^{2.4 } \frac{\(2.4)^2 }{2! }

p(X= 2) = e^{2.4 } \frac{\(2.4)^2 }{2! }= 0.261

b) The probability of getting more than '2'

P(X>2) = 1- {p(x=0)+p(x=1)+p(x=2)}

                = e^{-2.4 } \frac{(2.4)\^0 }{0! }+ e^{-2.4 } \frac{(2.4)\^1 }{1!}+e^{-2.4 } \frac{(2.4)\^2 }{2!}

              = 0.090 + 0.2177+0.261 = 0.566

P(x>2) = 0.566

c) The probability of getting between  two and five

P( 2<x<5) = P(x=3)+p(x=4) == e^{-2.4 } \frac{(2.4)\^3 }{3! }+ e^{-2.4 } \frac{(2.4)\^4 }{4!}+

P(2<x<5) = 0.2090 + 0.125 = 0.334

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