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2 years ago

A number cube is rolled. What is the probability that a number greater than 4 will be rolled?

Mathematics

2 answers:

ale4655 [162]2 years ago

7 0

Answer:

1/3

Step-by-step explanation:

Anything under 4 has a 4/6 chance, so whats left is 2/6, which is equal to 1/3.

slava [35]2 years ago

4 0

Answer:

1/3

Step-by-step explanation:

im assuming its a 6 sided dice

so each side is numbered 1-6

u want the probabilty if it being greater than 4

4 of the 6 numbers aren't greater than 4 so 4/6of the time it will be less than 4

which and be simplified to 2/3

so 1/1-2/3=1/3

so there is a 1/3 propbability that it will be greater than 4

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Given the function f(x)= √x+7, find the value of x when f(x)=1

solniwko [45]

F(x) = 1, so we make the entire equation equal 1.

1 = sqrt(x) + 7

To make x by itself subtract 7 from both sides

-6 = sqrt(x)

Square both sides to remove sqrt from x.

36 = x

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7 0

2 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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3 years ago

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tigry1 [53]

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tiny-mole [99]

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3 years ago