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Dmitry [639]
3 years ago
8

Ac=4 bx=y+4 what is the value of y

Mathematics
1 answer:
Kitty [74]3 years ago
6 0
Ac=4 bx=y+4 doesn't make sense.  Perhaps you meant <span>Ac-4bx=y+4.  If this is not correct, ensure that you have copied down the original problem correctly.

</span>To solve  Ac-4bx=y+4  for y, subtract 4 from both sides of this equation.  You'll get:

Ac-4bx-4 = y+4 - 4, or y = ac-4bx - 4.
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The U.S. National Highway Traffic Safety Administration gathers data concerning the causes of highway crashes where at least one
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Answer:

probability of a crash with at least one fatality if a driver drives while legally intoxicated (BAC greater than 0.09) = 0.001932

Step-by-step explanation:

P(BAC=0|Crash with fatality)=0.625

P(BAC is between .01 and .09|Crash with fatality)=0.302

P(BAC is greater than .09|Crash with fatality)=0.069

Let the event of BAC = 0 be X

Let the event of BAC between 0.01 and 0.09 be Y

Let the event of BAC greater than 0.09 be Z

Let the event of a crash with at least one fatality = C

P(X|C) = 0.625

P(Y|C) = 0.302

P(Z|C) = 0.069

P(C) = 0.028

probability of a crash with at least one fatality if a driver drives while legally intoxicated (BAC greater than 0.09) = P(C n Z)

But note that the conditional probability of probability that a driver is intoxicated (BAC greater than 0.09) given that there was a crash that involved at least a fatality is given by

P(Z|C) = P(Z n C)/P(C)

P(Z n C) = P(Z|C) × P(C) = 0.069 × 0.028 = 0.001932

Hope this Helps!!!

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Step-by-step explanation:

Just work it out

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Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ​ma
SOVA2 [1]

Answer:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X=10 represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Solution to the problem

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The sample mean \bar X is distributed on this way:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})  

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

The margin of error is given by:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

4 0
3 years ago
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