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solong [7]
3 years ago
13

Solve for p.

Mathematics
2 answers:
Marat540 [252]3 years ago
8 0
Hi there

3(p + q) = q
Solve for P

Use the distributive property
(3)(p) + (3)(q) = q
3p + 3q = q
3p = q-3q
3p = -2q
p = -2q/3

The answer is D

If you have any further questions please let me know :)
Andrej [43]3 years ago
8 0
3(p+q) = p

First we want to break apart the brackets. The 3 can be distributed as a coefficient to p and q .

3p + 3q = 1p

Next we want to isolate the variable p to a single side of the equation

3p + 3q - 3p = 1p - 3p

Simplifying the equation, we get

3q = -2p

Isolating for p, we get: 3q/-2 = p

This answer corresponds to answer D.
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Remember to show work and explain. Use the math font.
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Answer:

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Step-by-step explanation:

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2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------

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7 0
3 years ago
When 31 times a number is increased by 40 the answer is the same as two 200 is decreased by the number
storchak [24]
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4 0
3 years ago
Becky is competing in a 8-mi road race. She runs at a constant speed of 6 mi/h write an equation in slope intercept form to repr
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Answer:

y=-6x+8

Step-by-step explanation:

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6 0
3 years ago
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

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Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

6 0
3 years ago
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