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denis23 [38]
3 years ago
9

6. Solve for x. 10x - 4 = +(8x + 14)

Mathematics
2 answers:
Nuetrik [128]3 years ago
3 0

Simplify brackets

10x - 4 = 8x + 14

Add 4 to both sides

10x = 8x + 14 + 4

Simplify 8x + 14 + 4 to 8x + 18

10x = 8x + 18

Subtract 8x from both sides

10x - 8x = 18

Simplify 10x - 8x to 2x

2x = 18

Divide both sides by 2

x = 18/2

Simplify 18/2 to 9

<u>x = 9</u>

marshall27 [118]3 years ago
3 0

10x - 4 = 8x + 14

Step 1: Combine like terms

x's go with x's (10x and 8x). To do this subtract 8x to both sides

(10x - 8x) - 4 = (8x-8x) + 14

2x - 4 = 14

Normal numbers go with normal numbers (-4 and 14). To do this add 4 to both sides

2x + (-4 + 4) = 14 + 4

2x = 18

Step 2: Isolate x by dividing 2 to both sides

2x/2 = 18/2

x = 9

Hope this helped!

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Y = -27x + 577, where x represents the number of weeks and y represents the amount of money left in her account. After how many
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6 0
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Read 2 more answers
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makkiz [27]

Answer:

4 1/2

Step-by-step explanation:

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or

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8 0
2 years ago
The J.O. Supplies Company buys calculators from a non-US supplier. The probability of a defective calculator is 10 percent. If 3
RSB [31]

Answer:

There is a 24.3% probability that one of the calculators will be defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability of a defective calculator is 10 percent.

This means that p = 0.1

If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

This is P(X = 1) when n = 3. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.1)^{3}.(0.9)^{2} = 0.243

There is a 24.3% probability that one of the calculators will be defective.

3 0
3 years ago
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