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Gnesinka [82]
3 years ago
8

The door code to gain access to a top-secret laboratory is 6 digits. The first 3 digits of the code are all odd, and the last 3

digits are all even. Digits can be used more than once. How many possible codes are there?
Mathematics
1 answer:
WINSTONCH [101]3 years ago
3 0
5 odd numbers and 5 even numbers
since they can be repeated
5*5*5=first 3 digits
5*5*5=last 3 digits
in all
5*5*5*5*5*5=15625 ways
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both problems are proportional


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The selling price of an item is $600. It is marked down by 10%, but this sale price is still marked up
Lena [83]

Step-by-step explanation:

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= $540.

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6 0
3 years ago
Please help me with this question!!!!!
amm1812

Answer:

B. H(x)=-.008(x-82)^{2}+83

Step-by-step explanation:

Well lets graph the following equation first,

H(x) = -.011(x - 82)^2 + 75

Then lets graph all the other equations.

Look at the image below↓

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so A. is incorrect

B.

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C.

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3 0
4 years ago
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.

8 0
4 years ago
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