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3 years ago

I REALLLLLY NEED HELP WITH THAT!!!! Plz help

Mathematics

1 answer:

Inga [223]3 years ago

8 0

Answer:

If it says (10a + 4): 40a + 16

If it says (10a - 4): 40a - 16

Step-by-step explanation:

If it's an equilateral quadrilateral, that means all sides are equal. So you would multiply the given side by the number of sides. Then simplify you equation.

I can't read it if it says (10a + 4) or (10a - 4) so i'll just do both and you can use which one it says.

If it says (10a + 4):

4(10a + 4)

40a + 16

If it says (10a - 4)

4(10a - 4)

40a - 16

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A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

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Answer:

(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

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

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ 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, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

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

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

(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,

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

$\Rightarrow P(F)=(1-0.9)+(1-0.8)+(1-0.7)=0.6$

Lef $F_2$ be the event that she failed the 2nd exam, so

$P(F_2)=1-\frac{P(E_2)}{P(E_1)}$

$\Rightarrow P(F_2)=1-0.8=0.2$

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

$P\left(\frac{F_2}{F}\right)=\frac{P(F_2)}{P(F)}$

$\Rightarrow P\left(\frc{F_2}{F}\right)=\frac{0.2}{0.6}=\frac{1}{3}=0.33$

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