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Sloan [31]
4 years ago
10

Solve for c in this equation a(b-c)=a/b

Mathematics
2 answers:
Tpy6a [65]4 years ago
5 0

The problem is asking us to isolate c. This means that we are going to want to get c on its own side, with no other variables or terms. We can see that there is both an a and b on the side of the equation with c, so our goal will be to remove those.


To get rid of the a, let's divide both sides of the equation by a. This gives us:

\dfrac{a (b - c)}{a} = \dfrac{\frac{a}{b}}{a}

(b - c) = \dfrac{a}{b} \cdot \dfrac{1}{a} = \dfrac{1}{b}

(b - c) = \dfrac{1}{b}


Now, we are still going to have to remove the b. To do this, let's subtract a b from both sides of the equation. This gives us:

-c = \dfrac{1}{b} - b


Finally, let's multiply both sides of the equation by -1. This will remove the negative sign before the c term on the left side. This gives us:

c = b - \dfrac{1}{b}


Our answer is \boxed{c = b - \dfrac{1}{b}}

morpeh [17]4 years ago
4 0

c=b-1/b

This is the answer

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Step-by-step explanation:

Hello!

The study variable is

X: Pipe length.

It is known that this variable has a normal distribution and that the distribution parameter varies depending on the process used to manufacture the pipes.

Process A: μ= 200cm δ= 0.5cm

Process B: μ=201cm δ= 1cm

Process C: μ=202cm δ= 1.5cm

Pipes with a length of 200cm or more will be accepted by the utility company (X≥200), but pipes with less than 200cm length will be rejected (X<200)

a. Using Process C, you need to calculate the probability that the pipe will be rejected, symbolically:

P(X<200)

Using the distribution data of process C you have to standardize the value:

P(Z<(200-202)/1.5)= P(Z<-1.33)= 0.09176

b. The requirements change, accepting any pipe between 199 and 202, you have to calculate the probabilities of the pipes being between those lengths using the three process:

Process A:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-200)/0.5)) - P(Z≤(199-200)/0.5))

P(Z≤4) - P(Z≤-2) = 1 - 0.02275 = 0.97725

The probability of the pipe being rejected is 0.02275

Process B:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-201)/1)) - P(Z≤(199-201)/1))

P(Z≤1) - P(Z≤-2) = 0.84134 - 0.02275 = 0.81859

The probability of the pipe being rejected is 1-0.81859= 0.18141

Process C:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-202)/1.5)) - P(Z≤(199-202)/1.5))

P(Z≤0) - P(Z≤-2) = 0.5 - 0.02275 = 0.47725

The probability of the pipe being rejected is 1-0.47725= 0.52275

The pipes manufactured using process A has fewer chances of being rejected.

c.

Process A costs $140

P(X≥200)= 1 - P(X<200)= 1 - P(Z<0)= 1 - 0.5= 0.5

Process B costs $160

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1)= 1 - 0.15866= 0.84134

Process C costs $177

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1.33)= 1 - 0.09176= 0.90824

If they where to make 100 pipes:

Using process A: 100*0.5= 50 pipes will be accepted, so they'll win 50*($200-$140)= $3000

Using process B: 100*0.84134= 84.134≅ 84 pipes will beaccepted, so they'll win 84*($200-$160)= $3360

Using the process C: 100*0.90824= 80.824≅ 90 pipes will be accepted, so they'll win 90*($200-$177)= $2070

As you can see, using process B will maximize the profits.

I hope it helps!

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