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Ad libitum [116K]
3 years ago
9

Solve for x. Show EACH step for full credit. Circle your final answer

Mathematics
2 answers:
Stolb23 [73]3 years ago
5 0

Answer:

x = 4

Step-by-step explanation:

Remember to follow PEMDAS. PEMDAS is the order of operation and equals:

Parenthesis

Exponents (& Roots)

Multiplication

Division

Addition

Subtraction

~

First, distribute 2 to all terms within the parenthesis:

2(8 - 12x) = (2 * 8) - (2 * 12x) = 16 - 24x

Next, combine like terms. Like terms are terms with the same variables and the same amount of variables:

16 - 24x + 8x = -25x + 52

16 - 16x = -25x + 52

Isolate the variable, x. Add 25x and subtract 16 from both sides:

16 (-16) - 16x (+25x) = -25x (+25x) + 52 (-16)

-16x + 25x = 52 - 16

Combine like terms:

25x - 16x = 52 - 16

9x = 36

Isolate the variable, x. Divide 9 from both sides:

(9x)/9 = (36)/9

x = 36/9

x = 4

x = 4 is your final answer.

~

adelina 88 [10]3 years ago
3 0

Answer:

x=4

Step-by-step explanation:

2(8 - 12x) + 8x = -25x + 52

Distribute

16 -24x +8x = -25x +52

Combine like terms

16 - 16x = -25x +52

Add 25x to each side

16 -16x+25x = -25x+52+25x

16+9x = 52

Subtract 16 from each side

16+9x-16 = 52-16

9x = 36

Divide by 9

9x/9 = 36/9

x =4

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3 years ago
What percentage of babies born in the United States are classified as having a low birthweight (<2500g)? explain how you got
lawyer [7]

Answer:

2.28% of babies born in the United States having a low birth weight.

Step-by-step explanation:

<u>The complete question is</u>: In the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g. What percent of babies born in the United States are classified as having a low birth weight (< 2,500 g)? Explain how you got your answer.

We are given that in the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g.

Let X = <u><em>birth weights of newborn babies</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 3,500 g

            \sigma = standard deviation = 500 g

So, X ~ N(\mu=3500, \sigma^{2} = 500)

Now, the percent of babies born in the United States having a low birth weight is given by = P(X < 2500 mg)

         

   P(X < 2500 mg) = P( \frac{X-\mu}{\sigma} < \frac{2500-3500}{500} ) = P(Z < -2) = 1 - P(Z \leq 2)

                                                                 = 1 - 0.97725 = 0.02275 or 2.28%

The above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.97725.

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4 years ago
Answer for a follow...
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4 years ago
What I the slope of the line?<br><br> y-3=5(x-2)
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3 years ago
Read 2 more answers
HELP<br>8 Prove that 8^6+4^11 is divisible by 17. Answer __*17
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Answer:

<u>262,144</u> goes in the blank

Step-by-step explanation:

8^6 = 262,144

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4,456,448 ÷ 17 = 262,144

8 0
3 years ago
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