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EleoNora [17]
2 years ago
10

2(5x+4)=10x+6 Someone can help me?

Mathematics
2 answers:
Oliga [24]2 years ago
4 0

Answer:

<h2><em><u>HOP</u></em><em><u>E</u></em><em><u> THAT WILL</u></em><em><u> HELP</u></em><em><u> YOU</u></em></h2>

LenaWriter [7]2 years ago
3 0

Answer:

te best answer is

=-2

Step-by-step explanation:

espero que te ayude

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Step-by-step explanation:

0: -2

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3 years ago
30% increase on $50<br> help I need the answer nowww
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Answer:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

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2 years ago
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Which statement best describes the relationship between x and y in the equation y = 6 + x?
mash [69]

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A. The Value of y is six less than the value of x

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7 0
3 years ago
Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an lg sm
GuDViN [60]
Answer: 0.951%

Explanation:

Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.

Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by

P(x = k) = \frac{n!}{k!(n-k)!}p^k (1-p)^{n-k}

Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones). 

Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by

P(x \ \textless \  3) = P(x = 0) + P(x = 1) + P(x = 2)&#10;\\ \indent = \frac{12!}{0!(12-0)!}(0.54)^0 (1-0.54)^{12-0} + \frac{12!}{1!(12-1)!}(0.54)^1 (1-0.54)^{12-1} + \\ \indent \frac{12!}{2!(12-2)!}(0.54)^2 (1-0.54)^{12-2}&#10;\\&#10;\\ \indent = \frac{12!}{(1)(12!)}(0.46)^{12} + \frac{12(11!)}{(1)(11!)}(0.54)(0.46)^{11}+ \frac{12(11)(10!)}{(2)(10!)}(0.54)^2(0.46)^{10}&#10;\\&#10;\\ \indent = (1)(0.46)^{12} + (12)(0.54)(0.46)^{11}+ (66)(0.54)^2(0.46)^{10}&#10;\\ \indent \boxed{P(x \ \textless \  3) \approx 0.00951836732 }&#10;

Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.


5 0
3 years ago
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