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Sidana [21]
3 years ago
11

Events M and N are independent events. In this scenario, if P(M) = 0.46 and P(M and N) = 0.138, then P(N) = .

Mathematics
2 answers:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

P(N)=0.3

Step-by-step explanation:

Given: P(M)= 0.46, P(M and N)=0.138

Using P(M) ×P(N)= P(M and N)

⇒0.46×P(N)= 0.138

⇒P(N)= \frac{0.138}{0.46}

⇒P(N)=0.3

olganol [36]3 years ago
3 0

Answer:

The value of P(N) is, 0.3

Step-by-step explanation:

If two events i.e A and B are independent then;

P(A \cap B) = P(A) \cdot P(B)

As per the statement:

Events M and N are independent events.

If P(M) = 0.46 and P(M \cap N) = 0.138

To find the value of P(N):

By definition we have;

P(M \cap N) = P(M) \cdot P(N)

Substitute the given values we have;

0.138 = 0.46 \cdot P(N)

Divide both sides by 0.46 we have;

0.3 = P(N)

or

P(N) = 0.3

Therefore, the value of P(N) is, 0.3

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To find the population of bacteria 24 hours from now, we need to find the population of bacteria after every 8 hours.

The present population of the bacteria is 9315.

After 8 hours, the bacteria becomes double. So, the number of bacteria becomes 9315 x 2 = 18630.

Again after 8 hours, the bacteria becomes 18630 x 2 = 37260.

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Thus, after 24 hours from now, the population of the bacteria is 74520.

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1 year ago
Brianna has 0.15 liter of a liquid. She will use 0.4 of this liquid. How many liters will she use?
alukav5142 [94]

Answer: 0.0375

Step-by-step explanation:

0.15 divided by 4 = 0.0375

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3 years ago
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2 years ago
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

Step-by-step explanation:

Previous concepts

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".

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".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

6 0
2 years ago
A package of cookies, packed , ate the fourth part at breakfast , then 3/5 at tea time. there are still 18 cookies in the pakage
Leno4ka [110]
Ok so
1/4 of total eaten
3/4 left
3/5 of 3/4 eaten
9/20 left
18=9/20 of whole
multiply both sides by 20
360=9 of whole
divide both sides by 9
40=whole


40 total originlly

5 0
3 years ago
Read 2 more answers
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