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zimovet [89]
3 years ago
7

Please explain how it’s C?!

Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0
The equation of a circle centered at the origin is x² + y² = r², where r is the radius.

Because we know that the circle goes through point (0, -9), we know that the radius of the circle is 9, so the equation of the circle is x² + y² = 9² = x² + y² = 81.

Therefore, to see if the point (8, √17) lies on the circle, we need to substitute these coordinates into the equation:

x² + y² = 81
8² + (√17)² = 81
64 + 17 = 81
81 = 81

So this confirms that the point (8, √17) lies on the circle.

I hope this helps!
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Write forty-eight out as a hundred as a hundred as a fraction in simplest form.
slamgirl [31]
The original answer should be 48/100.
Reduced is 24/50
Even more is 12/25
Your answer is 12/25
5 0
3 years ago
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2m+5=5(m-7)-3m equals??
Svetradugi [14.3K]
2m+5=5(m-7)-3m
first, use distributive property...
2m+5=5m-35-3m
next, combine like terms on the right side...
2m+5=2m-35
next, subtract 2m from both sides.
2m+5=2m-35
-2m   -2m
you will get....
5=-35 in which is not true. This means the equation has no solution.
6 0
3 years ago
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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

7 0
2 years ago
Help me please i cant seem to get it
givi [52]

The diagnonals of a parallelogram bisect each other ,

so

=》 BD = 3x × 2

=》BD = 6x

and there's given that

=》

BD = \dfrac{3}{5} AC

=》

6x =  \dfrac{3}{5}  \times 40

=》

x =  \dfrac{24}{6}

=》

x = 4

so, if x = 4 , then the given quadrilateral is a parallelogram.

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kvv77 [185]
It is11 because 5 plus 6 is 11
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