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2 years ago

Help if u know the answer

Mathematics

1 answer:

Fiesta28 [93]2 years ago

6 0

$( \frac{a}{b} )( \frac{c}{d} )= \frac{ac}{bd}$ so
$( \frac{3}{10} )( \frac{-4}{5} )= \frac{3*-4}{10*5}= \frac{-12}{50}= \frac{-6}{25}$

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Light bulbs manufactured at a certain factory have a 3% probability of being defective. what is the probability that 5 out of a

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A common discrete distribution is used in statistics, as opposed to a continuous distribution is called a Binomial distribution. The probability that 5 out of a shipment of 30 will be defective is 0.001617.

<h3>What is Binomial distribution?</h3>

A common discrete distribution is used in statistics, as opposed to a continuous distribution is called a Binomial distribution. It is given by the formula,

P(x) = ⁿCₓ pˣ q⁽ⁿ⁻ˣ⁾

Where,

x is the number of successes needed,

n is the number of trials or sample size,

p is the probability of a single success, and

q is the probability of a single failure.

The probability of the bulb being defective is 3%(0.03) and the probability of the bulb not being defective is 97%(0.97).

Now, it is needed to be found that the probability that 5 out of a shipment of 30 will be defective can be written as,

P(x) = ⁿCₓ pˣ q⁽ⁿ⁻ˣ⁾

P(x=5) = ³⁰C₅ (0.03)⁵ (0.97)²⁵

P(x=5) = 0.001617

Hence, the probability that 5 out of a shipment of 30 will be defective is 0.001617.

Learn more about Binomial Distribution:

brainly.com/question/14565246

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Huw is paid a weekly wage. Every week he:

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if i have 12 apples, and then sally takes one, and billy takes 8, then i go pick 18 more apple but i eat 14, how many apples do

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A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th

umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable X is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, p = 0.09.

An item is defective irrespective of the others.

The random variable X follows a Binomial distribution with parameters n = 9 and p = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of X as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of X is 0.36.

Compute the variance of X as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of X is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

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2 years ago