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Hitman42 [59]
4 years ago
14

58% of what is 63.4​

Mathematics
2 answers:
zaharov [31]4 years ago
4 0

it is 36.772

if you want the work I can write it to!

Natali [406]4 years ago
3 0

Answer:

x=36.772

Step-by-step explanation:

1. We assume, that the number 63.4 is 100% - because it's the output value of the task.

2. We assume, that x is the value we are looking for.

3. If 63.4 is 100%, so we can write it down as 63.4=100%.

4. We know, that x is 58% of the output value, so we can write it down as x=58%.

5. Now we have two simple equations:

1) 63.4=100%

2) x=58%

where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:

63.4/x=100%/58%

6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for what is 58% of 63.4

63.4/x=100/58

(63.4/x)*x=(100/58)*x       - we multiply both sides of the equation by x

63.4=1.72413793103*x       - we divide both sides of the equation by (1.72413793103) to get x

63.4/1.72413793103=x

36.772=x

x=36.772

now we have:

58% of 63.4=36.772

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In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence
Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

\hat p =\frac{12}{300}=0.04

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

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{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.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

6 0
3 years ago
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Answer:

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Step-by-step explanation:

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