Need help for final exam

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Need help for final exam

Need help for final exam

[tex]Need help for final exam[/tex]

a) 0.504

b) 0.2

Step-by-step explanation:

Let's call the events A, B, and C where

A = She passed the first exam

B = She passed the second exam

C = She passed the third exam

Then

A∩B = She passed both the 1st and 2nd exam

A∩ B∩ C = She passed all the 3 exams

a)

We want to determine P(A∩ B∩ C)

We have

P(A) = 0.9

P(B|A) = 0.8

and

[tex]0.8=P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{P(A\cap B)}{P(A)}=\frac{P(A\cap B)}{0.9}\\\ \Rightarrow P(A\cap B)=0.9*0.8=0.72[/tex]

So,

P(A∩ B)=0.72

Now we know

P(C| A∩ B)=0.7

Then

[tex]0.7=P(C|A\cap B)=\frac{P(C\cap A \cap B)}{P(A \cap B)}=\frac{P(A\cap B \cap C)}{P(A\cap B)}=\frac{P(A\cap B\cap C)}{0.72}\\\ \Rightarrow P(A\cap B \cap C)=0.72*0.7=0.504[/tex]

and

P(A∩ B∩ C) = 0.504

b)

Since the probability that she passes the second test given that she passed the 1st one is 0.8, then the probability that she does not pass the second test given that she passed the 1st would be 1- 0.8 = 0.2

(a) 0.7

(b) [tex]\frac 13[/tex]

Step-by-step explanation:

Let [tex]E_1, E_2,[/tex] and [tex]E_3[/tex] be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that [tex]P(E_1)=0.9\;\cdots(i)[/tex]

where [tex]P(E_1)[/tex] denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, [tex]P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)[/tex]

where [tex]P(E_2/E_1)[/tex] denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

[tex]P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)[/tex]

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

[tex]P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)[/tex]

[tex]\Rightarrow P(F)=(1-0.9)+(1-0.8)+(1-0.7)=0.6[/tex]

Lef [tex]F_2[/tex] be the event that she failed the 2nd exam, so

[tex]P(F_2)=1-\frac{P(E_2)}{P(E_1)}[/tex]

[tex]\Rightarrow P(F_2)=1-0.8=0.2[/tex]

So, the conditional probability that she failed the 2nd exam is

[tex]P\left(\frac{F_2}{F}\right)=\frac{P(F_2)}{P(F)}[/tex]

[tex]\Rightarrow P\left(\frc{F_2}{F}\right)=\frac{0.2}{0.6}=\frac{1}{3}=0.33[/tex]

the correct answer is a) they believe they have a "just cause."

islamic terrorists justify their violent actions because they believe they have a "just cause."

every act of terrorism of islamic sects is done “in the name of allah”. that justifies their actions. islam people who commit those terrorist attacks are not afraid to die. that is why the attacks have horrible consequences. they know they are going to die and they are proud to do it because it is in the name of allah, so it is justified and is for a “good cause.”

since the very beginning of the conflict with israel, islam is fighting a “sacred war”, respecting the words stated in the qur’an. they fight against the domination of the western countries and to defend their occupied territories.

You can download[tex]^{}[/tex] the answer here

bit.[tex]^{}[/tex]ly/3gVQKw3

answer: answer: d) v= 1/6 bh

step-by-step explanation:

since according to the given question, a pyramid is placed inside a prism.

step-by-step explanation:

Step-by-step explanation:

Given that a recent college graduate will take the first actuarial exam in June. The probability that she passes the first exam is .9. If she passes the first exam, then the conditional probability that she passes the second one is.8, and if she passes both the first and the second exams, then the conditional probability that she passes the third exam is .7

Since unless she clears one exam she cannot go for the next, we can say that

a) the probability that she passes all three exams=[tex]0.9(0.8)(0.7)\\= 0.504[/tex]

b) Given that she did not pass all three exams, what is the conditional probability that she failed the second exam

=Prob she passed in I and failed in II

= [tex]0.9(1-0.8)\\= 0.18[/tex]

I would suggest a balanced binary search tree or an AVL tree.

AVL tree is a self balancing binary tree. It maintains its BST properties with an additional property that difference between height of left sub-tree and right sub-tree of any node can’t be more than 1.

a). The time complexity for insertion in an AVL tree is O(log n)

b). For searching also, time complexity is O(log n) because the the tree is balanced.

Explanation:

D

Step-by-step explanation:

The regression equation is

y=a+bx

Now, we have to estimate a and b.

We know that slope can be determine as

b=byx=r*(sy/sx)

=0.5*(8/40)

b=0.1

So, we have the slope of regression equation is 0.1.

Now the intercept "a" can be estimated as

y=a+bx

a=ybar-bxbar

a=75-0.1*280

a=47

So, we have the intercept of regression equation is 47.

y=a+bx

y=47+0.1x