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Crazy boy [7]
3 years ago
10

2 1/9÷2/3 help plsxxxxxxxxxxxxxxxz

Mathematics
2 answers:
vova2212 [387]3 years ago
5 0
The answer to your question is 3.666
Sauron [17]3 years ago
4 0

Answer:

19/6

Step-by-step explanation:

multiply the denominator 9 and 2, you'll get 18 then add 1. Multiply 2/3 by 3 to get 9/6. 19/9 times 9/6  = 171/54. Divide that fraction by 9 and you get 19/6.

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The six Grade is holding a fundraiser were students will make and sell peace necklaces for $1.50 each each student is responsibl
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$ 584.89

Step-by-step explanation:

so each student is responsible for selling 12 necklaces...and there are 130 students....so thats (130 * 12) = 1560 necklaces

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profit = revenue earned - expenses

profit = 2340 - 0.45 = 2339.55

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1/4 * 2339.55 = 2339.55/4 = 584.887 rounds to $ 584.89 <==

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3 years ago
At a sale this week, a table is being sold for $201 . This is a 33% discount from the original price. What is the original price
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The amount of revenue in dollars, y, that Amahd receives from selling x posters is given by the equation y = 4x. The cost of pro
n200080 [17]

Answer: B: 20

Step-by-step explanation:

You are given two functions that are both equal to y. What you would do is set each function equal to y and solve:

4x=3/2x + 50

Next you would subtract 3/2x to isolate the x to one side:

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Then you would multiply by 2/5 to get rid of the fraction so all you’re left with is x=__

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Hope this helps!

4 0
3 years ago
In a sample of nequals16 lichen​ specimens, the researchers found the mean and standard deviation of the amount of the radioacti
GuDViN [60]

Answer:

(a) The confidence level desired by the researchers is 95%.

(b) The sampling error is 0.002 microcurie per millilitre.

(c) The sample size necessary to obtain the desired estimate is 25.

Step-by-step explanation:

The mean and standard deviation of the amount of the radioactive​ element, cesium-137 present in a sample of <em>n</em> = 16 lichen specimen are:

\bar x=0.009\\s=0.005

Now it is provided that the researchers want to increase the sample size in order to estimate the mean μ to within 0.002 microcurie per millilitre of its true​ value, using a​ 95% confidence interval.

The (1 - <em>α</em>)% confidence interval for population mean (μ) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{s}{\sqrt{n}}

(a)

The confidence level is the probability that a particular value of the parameter under study falls within a specific interval of values.

In this case the researches wants to estimate the mean using the 95% confidence interval.

Thus, the confidence level desired by the researchers is 95%.

(b)

In case of statistical analysis, during the computation of a certain statistic, to estimate the value of the parameter under study, certain error occurs which are known as the sampling error.

In case of the estimate of parameter using a confidence interval the sampling error is known as the margin of error.

In this case the margin of error is 0.002 microcurie per millilitre.

(c)

The margin of error is computed using the formula:

MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table.

MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}

 0.002=1.96\times \frac{0.005}{\sqrt{n}}

       n=[\frac{1.96\times 0.005}{0.002}]^{2}

          =(4.9)^{2}\\=24.01\\\approx 25

Thus, the sample size necessary to obtain the desired estimate is 25.

6 0
4 years ago
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