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2 years ago

If (a^3+27)=(a+3)(a^2+ma+9) then m equals

Mathematics

1 answer:

Ronch [10]2 years ago

7 0

Answer:

m = - 3

Step-by-step explanation:

a³ + 27 ← is a sum of cubes and factors in general as

a³ + b³ = (a + b)(a² - ab + b²), thus

a³ + 27

= a³ + 3³

= (a + 3)(a² - 3a + 9)

comparing a² - 3a + 9 to a² + ma + 9, then

m = - 3

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2 years ago

Before the distribution of certain statistical software, every fourth compact disk (CD) is tested for accuracy. The testing proc

lilavasa [31]

Answer:

a) For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

$P(R) =\frac{1}{4} =0.25$

The failure probability for each program are given by:

$P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01$

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

$P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932$

So then the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

And if we use the fact that we have 4 possible test the true probability of interest would be:

$P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017$

b) $p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))$

And replacing we got:

$p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484$

c) From part a we now that the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

So then if we have 100 CDs the expected number of rejected Cd's are:

$100*0.068= 6.8 \approx 7$

Step-by-step explanation:

Part a

For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

$P(R) =\frac{1}{4} =0.25$

The failure probability for each program are given by:

$P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01$

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

$P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932$

So then the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

And if we use the fact that we have 4 possible test the true probability of interest would be:

$P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017$

Part b

For this case we want the probability that it failed program 2 or 3

So then we can find this probability like this:

$p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))$

And replacing we got:

$p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484$

Part c

From part a we now that the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

So then if we have 100 CDs the expected number of rejected Cd's are:

$100*0.068= 6.8 \approx 7$

3 0

2 years ago