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Vesna [10]
2 years ago
7

P= s-c/s        solve for c

Mathematics
2 answers:
Leokris [45]2 years ago
6 0
p= s - \frac{c}{s}
p=  \frac{ s^{2} }{s} -  \frac{c}{s}
p= \frac{s^2-c}{s}
ps= s^2-c
ps + c = s^2
c =s^2-ps


Natalka [10]2 years ago
4 0

Answer:

c = s (s-P)

Step-by-step explanation:

To solve the expression for c, we have to do different operations in order to leave c on one side and the rest of the variables on the other.

First we have:

P= s-\frac{c}{s}

The next step would make the minimum common multiple for the denominator: P= \frac{s}{1} -\frac{c}{s} \\P=\frac{s^{2}-c }{s}

Now we're going to cross multiply:

P=\frac{s^{2}-c }{s}\\Ps=s^{2} -c

Now we're going to move the s² to the other side:

\\Ps-s^{2} =-c

We have -c instead of c, so we're going to change the signs of all terms:

-Ps+s^{2} =c

Now we solved for c but we can still make one more step because both terms on the left side have an s.

-Ps+s^{2} =c\\s(-P+s) = c

Thus c = s (s-P)

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Answer:

Jillian is incorrect because square roots of perfect squares are rational.

Step-by-step explanation:

√25 = √(5²) = 5 - rational

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6 0
2 years ago
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Player V and Player M have competed against each other many times. Historical data show that each player is equally likely to wi
Helen [10]

Answer:

0.46 (46%)

Step-by-step explanation:

We have the following data:

- Probability that player V wins the first set:

p(1)=0.50

(because the text says the two players are equally likely to win the first set)

- Probability that player V wins the 2nd set if he has won the 1st set:

p(2|1)=0.60

So, the probability that player V wins the first 2 sets is:

p(12)=p(1)\cdot p(2|1)=(0.50)(0.60)=0.30 (1)

Instead, the probability that player V loses the 2nd set if he has won the 1st set is 0.40 (=1-0.60), so

p(2^c|1)=0.40

So, the probabiity that player V winse the 1st set but loses the 2nd set is

p(12^c)=p(1)\cdot p(2^c|1)=(0.50)(0.40)=0.20 (2)

Also, we have:

- Probability that player V loses the 1st set:

p(1^c)=0.50

- Probability that she will lose the 2nd set in this case is 0.70, it means that the probability that she will win the 2nd set if she lost the 1st set is 0.30, so:

p(2|1^c)=0.30

So, the probability that she will lose the 1st set and win the 2nd set is:

p(1^c2)=p(1^c)\cdot p(2|1^c)=(0.50)(0.30)=0.15 (3)

Combined together (2) and (3), this means that the probability that player V wins exactly 1 set out of the first two sets is:

p(1/2)=p(12^c)+p(1^c2)=0.20+0.15=0.35 (4)

At this point, the probability that she will win the 3rd set is

p(3)=0.45

This means that the overall probability that she will win the 3rd set if she won exacty 1 of the first 2 sets is:

p(1/2,3) = p(1/2)\cdot p(3)=(0.35)(0.45)=0.16 (5)

So, the overall probability that player V will win a match against player M is the sum of (1) and (5):

p(W)=p(12)+p(1/2,3)=0.30+0.16=0.46

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2 years ago
Roger manages an ice cream stand. He keeps track of how many minutes each customer stands in line.
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Answer:

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