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katrin [286]
3 years ago
12

The system of equations below represents p, the number of photo files, and v, the number of video files Helena uploaded to an on

line storage account. p + v = 650 4p + 9v = 2,925 Which statement describes the solution to the system of equations? Helena uploaded 520 more video files than photo files. Helena uploaded 65 fewer video files than photo files. Helena uploaded 9 times as many photo files as video files.
Mathematics
2 answers:
ss7ja [257]3 years ago
5 0
P + v = 650....p = 650 - v
4p + 9v = 2925

4(650 - v) + 9v = 2925
2600 - 4v + 9v = 2925
-4v + 9v = 2925 - 2600
5v = 325
v = 325/5
v = 65.....video files

p + v = 650
p + 65 = 650
p = 650 - 65
p = 585....photo files

answer : Helena uploaded 9 times as many photo files as video files
               65 * 9 = 585
Trava [24]3 years ago
4 0

Answer: 9 times as many photo files as video files

Step-by-step explanation:

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15x + 2 = 45x + 4

Subtract 2 from both sides, as well as subtracting 45x from both sides.

15x - 45x = 4 - 2

-30x = 2

Divide both sides by -30.

x = 2/-30

x = - 1/15

~Hope I helped!~
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Answer:

d or 4x²-12x-9

Step-by-step explanation:

factoring d you get (2x-3)²

which is a perfect square trinomial so

d or 4x²-12x-9

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Lorico [155]

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The perimeter of a rectangle is 42 inches. If the width of the rectangle is 6 inches, what is the length?
snow_lady [41]

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8 0
3 years ago
Read 2 more answers
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

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

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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