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Volgvan
3 years ago
5

For positive test result, the number of those who did not lie and lie are 11 and 41, respectively. Those of the negative test re

sult are 30 and 8. Find P(subject lied I negative results) and P(negative results I subject lied).
Mathematics
1 answer:
vagabundo [1.1K]3 years ago
3 0

P(subject lied | negative results) = 4/19

P(negative results | subject lied) = 8/49

 

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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The temperature would be 55 degrees fahrenheit
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Solve for the equation 4a-8=28
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3 years ago
A welder drops a piece of red-hot steel on the floor. The initial temperature of the steel is 2,500 degrees Fahrenheit. The ambi
sergey [27]

Answer: After 18.05 minutes, the temperature of steel becomes 100 degrees.

Step-by-step explanation:

Since we have given that

Initial temperature = 2500

At t = 0,

we get that

f(t)=Ce^{-kt}+80\\\\2500=C+80\\\\2500-80=C\\\\2420=C

After 2 minutes, the temperature of the steel is 1500 degrees.

so, it becomes,

1500=2420e^{-2k}+80\\\\1500-80=2420e^{-2k}\\\\\dfrac{1420}{2420}=e^{-2k}\\\\0.587=e^{-2k}\\\\\ln 0.587=-2k\\\\-0.533=-2k\\\\k=\dfrac{0.533}{2}\\\\k=0.266

So, We need to find the number of minutes when the temperature of steel would be 100 degrees.

So, it becomes,

100=2420e^{-0.266t}+80\\\\100-80=2420e^{-0.266t}\\\\20=2420e^{-0.266t}\\\\\dfrac{20}{2420}=e^{-0.266t}\\\\\ln \dfrac{20}{2420}=-0.266t\\\\-4.8=-0.266t\\\\t=\dfrac{4.8}{0.266}\\\\t=18.05

Hence, after 18.05 minutes, the temperature of steel becomes 100 degrees.

7 0
3 years ago
In the third week of July, a random sample of 40 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. A
suter [353]

Answer:

a. [6.6350,7.3950]

b. ME=0.5150

Step-by-step explanation:

a. Given that n=40, \bar x=6.88 and that:z_{\alpha/2}=z_{0.05}=1.645

The required 90% confidence interval can be calculated as:

\bar x\pm(margin \ of \ error)\\\\\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\6.88\pm(1.645\times \frac{1.98}{\sqrt{40}})\\\\=[6.3650,7.3950]

Hence, the 90% confidence interval for the population mean cash value of this crop is [6.6350,7.3950]

b. The margin of error at 90% confidence interval is calculated as:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\=(1.645\times \frac{1.98}{\sqrt{40}})\\\\=0.5150

Hence, the margin of error is 0.5150

8 0
4 years ago
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