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4 years ago

Identify two values that have a value less than 3 what is the answer to this question?

Mathematics

2 answers:

Digiron [165]4 years ago

8 0

Answer:

the answer is 2.9 x 10^0 and 3.2 x 10^-2

Step-by-step explanation:

10^0 is equal to 1 so 2.9 x 10^0 is really 2.9 times 1 which is less than 3

if i could get brainliest answer that would be great!

10^-2 is a exponential that makes a number go down so 3.2 going down is less than 3

DerKrebs [107]4 years ago

8 0

Answer:

C and D

Step-by-step explanation:

A. 2.9 × 10¹ = 2.9 × 10 = 29

B. 3.2 × 10³ = 3.2 × 1000 = 3000

C. 2.9 × 10⁰ = 2.9 × 1 = 2.9

D. 3.2 × 10⁻² = 3.2 × 0.01 = 0.032

Both C and D have values less than 3.

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Answer:

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3 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) $COV = 0.007255$

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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4 years ago