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fiasKO [112]
3 years ago
11

3y-3.5/3y+6=17/25.5. find the value of y. give the scale factor of the polygon

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
3 0

Hi the answer to your question is

 

((3y - 3.5) / (3y + 6)) * 25.5 = (17 / 25.5) * 25.5

((3y - 3.5) / (3y + 6)) * 25.5 = 17 (76.5y - 89.25) / (3y + 6) = 17

((76.5y - 89.25) / (3y + 6)) * (3y + 6) = 17 * (3y + 6) 76.5y - 89.25 = 51y + 102 76.5y - 51y = 102 + 89.25

25.5y = 191.25

y = 191.25 / 25.5

<span>y = 7.5</span>

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URGENT, please!!
lakkis [162]

The equation of the parabola is:

y = 24x^2

---------------------------

The <em>equation </em>of a parabola of vertex (h,k) and focus (0,f) is given by:

y = 4p(x - h)^2 + k^2

  • Vertex at the origin means that h = 0, k = 0.
  • Focus (0,6) means that p = 6

Thus, the equation of the parabola is:

y = 4p(x - h)^2 + k^2

y = 4(6)(x - 0)^2 + 0^2

y = 24x^2

A similar problem is given at brainly.com/question/15165354

7 0
3 years ago
Suppose the following number of defects has been found in successive samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6
Brut [27]

Answer:

Given the data in the question;

Samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.

a)

For a p chart ( control chart for fraction nonconforming), the center line and upper and lower control limits are;

UCL = p" + 3√[ (p"(1-P")) / n ]

CL = p"

LCL = p" - 3√[ (p"(1-P")) / n ]

here, p" is the average fraction defective

Now, with the 30 samples of size 100

p" =  [∑(6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.)] / [ 30 × 100 ]

p" = 234 / 3000

p" = 0.078

so the trial control limits for the fraction-defective control chart are;

UCL = p" + 3√[ (p"(1-P")) / n ]

UCL = 0.078 + 3√[ (0.078(1-0.078)) / 100 ]

UCL = 0.078 + ( 3 × 0.026817 )

UCL = 0.078 + 0.080451

UCL = 0.1585

LCL = p" - 3√[ (p"(1-P")) / n ]

LCL = 0.078 - 3√[ (0.078(1-0.078)) / 100 ]

LCL = 0.078 - ( 3 × 0.026817 )

LCL = 0.078 - 0.080451

LCL =  0 ( SET TO ZERO )

Diagram of the Chart uploaded below

b)

from the p chart for a) below, sample 28 violated the first western electric rule,

summary report from Minitab;

TEST 1. One point more than 3.00 standard deviations from the center line.

Test failed at points: 28

Hence, we conclude that the process is out of statistical control

Lets Assume that assignable causes can be found to eliminate out of control points.

Since 28 is out of control, we should eliminate this sample and recalculate the trial control limits for the P chart.

so

p" = 0.0745

UCL = p" + 3√[ (p"(1-P")) / n ]

UCL = 0.0745 + 3√[ (0.0745(1-0.0745)) / 100 ]

UCL = 0.0745 + ( 3 × 0.026258 )

UCL = 0.0745 + 0.078774

UCL = 0.1532

LCL  = p" - 3√[ (p"(1-P")) / n ]

LCL = 0.0745 - 3√[ (0.0745(1-0.0745)) / 100 ]

LCL = 0.0745 - ( 3 × 0.026258 )

LCL = 0.0745 - 0.078774

UCL = 0  ( SET TO ZERO )

The second p chart diagram is upload below;

NOTE; the red circle symbol on 28 denotes that the point is not used in computing the control limits

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The graph of an equation with a negative discriminant always has which characteristic?<br>​
jasenka [17]

Answer:

a posative

Step-by-step explanation:

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3 years ago
The weather forecast states that the temperature will be lower than –5°F tonight. Let t represent the temperature in degrees Fah
Deffense [45]
Hi There!

<span>t < –5 and t may be a negative number!</span>
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Read 2 more answers
complete the puzzle by unscrambling the letters below to reveal words from the vocabulary list at the beginning of the chapter
Marina CMI [18]

But what are we meant to unscramble?

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