A fair coin is flipped four times. Use a tree diagram to find the probability of observing exactly two heads.

The nursing department of a college surveyed two hundred graduates from their programs about their current

mel-nik [20]

Answer: The required probability is 0.008.

Step-by-step explanation:

Since we have given that

Probability of bachelor's degree = 0.45

Probability of working in nursing = 0.85

Probability of both = 0.4

So, Probability of getting a graduate is currently working in nursing, given that they earned a bachelor's degree would be :

$P(N|B)=\dfrac{P(N\cap B)}{P(B)}\\\\P(N|B)=\dfrac{0.4}{0.45}\\\\P(N|B)=0.008$

Hence, the required probability is 0.008.

3 0

3 years ago

Which term below correctly completes the following sentence?

Norma-Jean [14]

Answer:

c

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Show that in any group of 101 people there are at least two persons having the same number friends (It is assumed that if a pers

vladimir1956 [14]

Answer:

Step-by-step explanation:

Let us assume that if possible in the group of 101, each had a different number of friends. Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.

Say P has 100 friends then P has all other persons as friends. In this case, there cannot be any one who has 0 friend. So a contradiction. Hence proved

Part ii: EVen if instead of 101, say n people are there same proof follows as

if different number of friends then they would be 0,1,2...n-1

If one person has n-1 friends then there cannot be any one who does not have any friend.

Thus same proof follows.

7 0

3 years ago

Find the midpoint of the segment with the given endpoints. (9,9) and (-1,- 3)

GenaCL600 [577]

Answer: (4, 3)

Step-by-step explanation:

The formula for coordinate of the mid point is given as :

Mid point = ( $\frac{x_{2}+x_{1}}{2}$ , $\frac{y_{2}+y_{1}}{2}$ )

$x_{1}$ = 9

$x_{2}$ = -1

$y_{1}$ = 9

$y_{2}$ = -3

Substituting the values into the formula , we have :

Mid-point = ( $\frac{-1+9}{2}$ , $\frac{-3+9}{2}$ )

Mid-point = ( $\frac{8}{2}$ , $\frac{6}{2}$ )

Mid - point = ( 4 , 3)

8 0

3 years ago

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar

Mamont248 [21]

Answer:

a

The 95% confidence interval is

$0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

c

$n = 125 \ stores$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The number of stores that had more than 2 items price incorrectly is k = 52

Generally the sample proportion is mathematically represented as

$\^ p = \frac{ k }{ n }$

=> $\^ p = \frac{ 52 }{ 60 }$

=> $\^ p = 0.867$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=> $E = 0.0859$

Generally 95% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.867 - 0.0859 < p < 0.867 + 0.0859$

=> $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Considering question b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

Considering question c

From the question we are told that

The margin of error is E = 0.05

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n= [\frac{1.645 }}{0.05} ]^2 * 0.867 (1 - 0.867 )$

=> $n = 125 \ stores$

4 0

3 years ago