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maria [59]
3 years ago
9

2p - 1/q = r/q + 4 , subject q

Mathematics
1 answer:
kondaur [170]3 years ago
4 0

Solve for q:

2 p - 1/q = r/q + 4

Bring 2 p - 1/q together using the common denominator q. Bring r/q + 4 together using the common denominator q:

(2 p q - 1)/q = (4 q + r)/q

Multiply both sides by q:

2 p q - 1 = 4 q + r

Subtract 4 q - 1 from both sides:

q (2 p - 4) = r + 1

Divide both sides by 2 p - 4:

Answer:  q = (r + 1)/(2 p - 4)

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Patti's Pet Palace has a 90-gallon fish tank. When it needs to be cleaned, it can be drained at a rate of 2 gallons per minute.
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Simplify: 6(5x - 2) - (3x - 4)​
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Step-by-step explanation:

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Karla makes paper airplanes. Today, it takes her 32 min to make 16 paper airplanes.
Pavel [41]

Answer:

48/b

Step-by-step explanation:

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7 0
3 years ago
An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

7 0
3 years ago
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