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lina2011 [118]
3 years ago
14

Space Shuttle has three computers. Computer A is a primary computer and Computer B and Computer C are auxiliary computers. There

is a constant probability of 3% that each computer might malfunction. What is the probability that during the operation of leaving orbit, primary computer fails, and one of the auxiliary computers will be in use?
Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
3 0

Answer:

Required probability equals 0.18%

Step-by-step explanation:

The probability that the primary and one auxiliary computer fails equals

1) Probability that A and B fails

2)Probability that A and C fails

Thus required probability equals

P(E)=P(1)(2)+P(1)P(3)\\\\P(E)=0.03\times 0.03+0.03\times 0.03P(E)=0.18%

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Pythagorean theory . Thank you
svp [43]

Answer:

b=12, perimeter=36

7 0
2 years ago
The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

N = N_0 * (\frac{1}{2})^{t/p

N_0 is the starting amount of parent element.

N is the end amount of parent element

t is the time elapsed

p is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

N(t) = 74g * (\frac{1}{2})^{t/2.8days

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2}  =  37g

c.

Now we just gotta do some algebra. Use the original function but this time, replace N(t) with 10g and solve algebraically.

10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}

Take the log of both sides.

log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})

Use the exponent rule for log laws that, log(b^x) = x*log(b)

log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})

\frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = \frac{t}{2.8days}

2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = t

slap that in your calculator and you get

t = 8.1 days

7 0
2 years ago
Which events are mutually exclusive?
statuscvo [17]
The last one is I think
6 0
3 years ago
A doghouse is to be built in the shape of a right trapezoid, as shown below. What is the area of the doghouse? top is 7 side 7 b
Ann [662]
I believe the answer is 45.5 square feet

8 0
3 years ago
A recent study of the relationship between social activity and education for a sample of corporate executives showed the followi
jekas [21]

Answer:

Chisquare statistic

Step-by-step explanation:

The most appropriate test to use here is the Chisquare test as it isemployed when testing the relationship between two variables. Both the T statistic and z statistics are the used for testing difference in means. However, in the analysis above, we are given two variables with each having its own levels. Hence, the Chisquare test is the most appropriate in this kind of situation a it measures if a difference exists between the two variables. It tests if they are related or independent of on one another

7 0
3 years ago
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