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Vilka [71]
3 years ago
5

The probability of event A is x, and the probability of event B is y. If the two events are independent, which of these conditio

ns must be true?

Mathematics
1 answer:
Inga [223]3 years ago
5 0

Answer:

P(B|A)=y

Step-by-step explanation:

The complete question is shown in the attachment.

If the events, A and B are independent, then

P(A and B) = P(A)×P(B)

But from the question:

P(A)=x and P(B)=y

This implies that:

P(A and B) = xy

Also, the conditional probabilities are:

P(A\B)=P(A)=x

P(B\A)=P(B)=y

Therefore the correct answer is A

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Is there a mistake in this equation? if not please let me know!
Juli2301 [7.4K]

Answer:

no mistakes

Step-by-step explanation:

4 * 3^x = 324

Divide each side by 4

4/4 * 3^x = 324/4

3^x = 81

Rewriting 81 as a power of 3

3^x = 3^4

Since the bases are the same, the powers are the same

x=4

5 0
3 years ago
Read 2 more answers
A^2+b^2+c^4=2020. Yes
pychu [463]

Answer:

yes

Step-by-step explanation:

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3 years ago
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16, 4,1,...<br> Find the 6th term
Vlada [557]

Let's see what to do...

_________________________________

Step (1)

Look : 16 , 10 , 4 , ...

It is an arithmetic sequence.

So first we have to find the value of the ratio(( d )) .

To do this, we need to subtract each sentence minus the previous sentence.

So we have : 10 - 16 = - 6

_________________________________

Step (2)

To find the nth sentence in an arithmetic sequence, we must use the following equation :

t(n) = t(1) + (n - 1)d

_________________________________

Step (3)

To find the 6th term just need to put the following parameters in the above sequence.

t(1) = 16 ------ n = 6 ------ d = -6

t(6) = 16 + (6 - 1)( - 6)

t(6) = 16 + (5)( - 6)

t(6) = 16 + ( - 30)

t(6) = 16 - 30

t(6) =  - 14

_________________________________

And we're done.

Thanks for watching buddy good luck.

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3 0
3 years ago
The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma
nata0808 [166]

Answer:

a) \sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b) Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

c) P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d) P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e) P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

X \sim N(110000,40000)  

Where \mu=110000 and \sigma=40000

If we select a sample size of n =35 the standard error is given by:

\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

P(X > 112000)

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

P(X > 100000)

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

P(100000

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

8 0
3 years ago
In a football league, 4 coaches are needed for every 38 players. If 84 players sign up for softball, how many coaches are needed
pickupchik [31]
9


8.842 but the answer would have to be rounded because you can’t have .842 coaches... the answers 9!!!!
6 0
3 years ago
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