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Firdavs [7]
3 years ago
12

The Environmental Protection Agency (EPA) has contracted with your company for equipment to monitor water quality for several la

kes in your water district. A total of 10 devices will be used. Assume that each device has a probability of 0.01 of failure during the course of the monitoring period. What is the probability that none of your devices fail
Mathematics
2 answers:
anastassius [24]3 years ago
7 0

The probability that none of the devices fails happens when x = 0.

P(none fail) = 1(0.01)^(0)•(1 - 0.01)^(10)

P(none fail) = (0.01)^0•(0.99)^(10)

P(none fail) = 0.90438

sergij07 [2.7K]3 years ago
7 0

Answer:

Probability that none of your devices fail is 0.9044.

Step-by-step explanation:

We are given that the Environmental Protection Agency (EPA) has contracted with your company for equipment to monitor water quality for several lakes in your water district. A total of 10 devices will be used. Assume that each device has a probability of 0.01 of failure during the course of the monitoring period.

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 10 devices

             r = number of success = none fail

            p = probability of success which is probability of failure during the

                   course of the monitoring period, i.e; 0.01.

<em>LET X = No. of failures</em>

So, it means X ~ Binom(n=10, p=0.01)

Now, Probability that none of your devices fail is given by = P(X = 0)

      P(X = 0) =  \binom{10}{0} \times 0.01^{0} \times (1-0.01)^{10-0}

                    = 1 \times 1 \times 0.99^{10} = 0.9044

Hence, the probability that none of your devices fail is 0.9044.

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Answer:

  • <em>convert </em>the mixed fractions to improper fractions (where the numerator is greater than or equal to the denominator): multiply the whole number part by the fraction's denominator, add that to the numerator, write the result on top of the denominator.
  • if the denominators are not the same, work out the common denominator and <em>rewrite </em>the fractions with the same denominators
  • subtract by subtracting the numerators and writing the result over the denominator
  • convert back to mixed fractions by dividing the numerator by the denominator, write down the whole number answer, write down the remainder above the denominator.

Example

3\frac23-1\frac45

convert to improper fractions:

\dfrac{3 \times 3+2}{3}-\dfrac{1 \times 5+4}{5}=\dfrac{11}{3}-\dfrac{9}{5}

common denominator = 3 × 5 = 15, so:

\dfrac{11}{3}-\dfrac{9}{5}=\dfrac{11\times 5}{3\times 5}-\dfrac{9\times 3}{5\times 3}=\dfrac{55}{15}-\dfrac{27}{15}

subtract:

\dfrac{55}{15}-\dfrac{27}{15}=\dfrac{55-27}{15}=\dfrac{28}{15}

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What are you trying to say?
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I believe you should try to plug in the ordered pair to see if the solutions work.

The ordered pair is (x,y) because x is on the x-axis and y is on the y-axis, therefore in (16,-3) the x=16 and y=-3.

Now substitute, meaning plug in those numbers into the equation.
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Now you just need to solve one side and see if it equals the other side.
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First you use order of operations to solve this. PEMDAS. So you multiply 2(-3) first because of the parenthesis and it being multiplication.
16+(-6) and when you have a positive number adding a negative number it’s going backwards of the number line, basically meaning subtraction in a way. Sorry if this confuses you, if you already know how to do negatives and such nevermind this part.
But 16+(-6)=10
So now you look at both side of the equation, does the left side equal to the right? 10=10, so yes. It is a solution for that equation.

Now for the next equation, 7y=-21
Again, plug in the ordered pair (16,-3) into the equation. Remember that it’s (x,y).
There is no x in this equation so no need to worry about that; you only plug in y for this one.
7(-3) Now you multiply. Whenever you multiply a positive number and a negative number, the answer will always be negative. So 7(-3) is -21.
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(16,-3) is a solution to both equations.

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