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lisabon 2012 [21]
2 years ago
7

Please help me with problem with clear explanation

Mathematics
2 answers:
Svet_ta [14]2 years ago
6 0

Answer:

this doesn't even make any sense

Step-by-step explanation:

both -5.4 and 5.4 go off the m.f grid

bezimeni [28]2 years ago
3 0
1. •The starting point will be located in quadrant Q
•The finishing point will be located in quadrant S
Explanation: the x and y values determine where the points are located on the graph

2. The starting point is (-5.4, 3) and the checkpoint is (5.4, 3). The difference is they’re both in different quadrants, and one has a negative value, and the other is positive. They are related because they’re all in one race, and the checkpoint will be involved in the race, most likely between the start and finish points, located in quadrant P.

I hope that helps, merry Christmas and happy holidays! :)
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S - 10,000 = 3,753<br><br>A. s = 13,573<br><br>B. s = 13,753<br><br>C. s = 10,375
Lesechka [4]

Answer:

B. s = 13753

Step-by-step explanation:

s - 10000 = 3753

To find value of s, add both sides by 10000

s - 10000 = 3753

   +10000    +10000

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

s = 13753

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3 years ago
Curt and Ian both ran a mile. Curt time was 8/9 of Ian time. Who ran faster?
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2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi
Oksana_A [137]

Answer:

(a) E(X) = 950

(b) $ COV = 0.007255$

(c) P(X > 980) = 0.00001\\\\

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$ COV = \frac{\sigma}{E(X)} $

Where the standard deviation is given by

\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892

So the coefficient of variance is

$ COV = \frac{6.892}{950} $

$ COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

P(X > 980)  = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980)  = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980)  = 1 - P(Z < 4.28)\\\\

The z-score corresponding to 4.28 is 0.99999

P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\

So it means that it is very unlikely that there will be more than 980 non-defective units.

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