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3 years ago

For which value of x does each expression make sense

Mathematics

1 answer:

Vitek1552 [10]3 years ago

6 0

$\text{If you want the domain of this function then it is}~x\leq 0 ~ \text{or} ~ (-\infty ,0]$

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3.16 x 10 with the power of 3

DENIUS [597]

Answer:

3160

Step-by-step explanation:

3.16 x 10^3

10^3 = 1000

so you move the decimal point three times to the right

I hope this helps

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Exponential growth and decay:

Stella [2.4K]

Answer:

125 g

Step-by-step explanation:

500/2=250

250/2=125

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3 years ago

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If EFG ≅ △ ABD<br> so FG is<br> = .......?<br> help asap

Igoryamba

Answer:

FG=BD

Step-by-step explanation:

hope this helps!!

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3 years ago

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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) $COV = 0.007255$

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

Using sq root property. 2x^2+36+0

mezya [45]

It’s 2.&:&/8/8/8&/keisjsisk

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3 years ago