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scoundrel [369]
3 years ago
14

If your gross pay for last month was $2,180, what is your net pay if your have no federal or state income tax withholdings?

Mathematics
1 answer:
dmitriy555 [2]3 years ago
8 0
Probably the same amount, booyakasha
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to what

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Sherita needs 1 3/5 cups of sugar. she has 8 cups of sugar. how much cups can she make?
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PLEASE HELP
Nikolay [14]

Answer:

204,160,000

Step-by-step explanation:

Assuming the sample is a good representation of the people from the United States.

Since there were 1,000 surveyed... and 638 of them were wearing glasses,  that makes a proportion of 638 out of  1,000 people, or 63,8 / 100 people... so 63.8%... or 0.638

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The quantity x minus 3 divided by y
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x-3/y

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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)
\\ \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}
\\
\\ \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}
\\
\\ \indent = (1)(0.46)^{12} + (12)(0.54)(0.46)^{11}+ (66)(0.54)^2(0.46)^{10}
\\ \indent \boxed{P(x \ \textless \  3) \approx 0.00951836732 }


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


5 0
4 years ago
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