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shtirl [24]
2 years ago
13

uppose you start at the origin, move along the x-axis a distance of 7 units in the positive direction, and then move downward al

ong the z-axis a distance of 6 units. What are the coordinates of your position
Mathematics
1 answer:
yan [13]2 years ago
6 0
(7, -6). This is because you are moving 7 to the right (in other words, ADDING 7 to the x value) and going DOWN 6 units on the y axis (subtracting from y value). Remember that the origin is (0, 0) and coordinates are represented as (x, y)
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3 years ago
Read 2 more answers
A company is reviewing a batch of 24 products to determine if any are defective. On average, 3.2% of products are defective.
Svetach [21]

Using the binomial distribution, it is found that:

  • 0.9599 = 95.99% probability that the company will find 2 or fewer defective products in this batch.
  • 0.0066 = 0.66% probability that 4 or more defective products are found in this batch.

-----------------

For each product, there are only two possible outcomes, either it is defective, or it is not. The probability of a product being defective is independent of any other product, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the<u> probability of exactly x successes on n repeated trials.</u>

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by:

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of a success on a single trial.

-----------------

  • 24 products means that n = 24
  • 3.2% are defective, thus p = 0.032

-----------------

The probability that <u>2 or fewer are defective</u> is:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{24,0}.(0.032)^{0}.(0.968)^{24} = 0.4582

P(X = 1) = C_{24,1}.(0.032)^{1}.(0.968)^{23} = 0.3635

P(X = 2) = C_{24,2}.(0.032)^{2}.(0.968)^{22} = 0.1382

Thus

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.4582 + 0.3635 + 0.1382 = 0.9599

0.9599 = 95.99% probability that the company will find 2 or fewer defective products in this batch.

-----------------

The probability that <u>4 or more are defective</u> is:

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Then

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{24,0}.(0.032)^{0}.(0.968)^{24} = 0.4582

P(X = 1) = C_{24,1}.(0.032)^{1}.(0.968)^{23} = 0.3635

P(X = 2) = C_{24,2}.(0.032)^{2}.(0.968)^{22} = 0.1382

P(X = 3) = C_{24,3}.(0.032)^{3}.(0.968)^{21} = 0.0335

Thus

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.4582 + 0.3635 + 0.1382 + 0.0335 = 0.9934

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.9934 = 0.0066

0.0066 = 0.66% probability that 4 or more defective products are found in this batch.

A similar problem is given at brainly.com/question/23780714

6 0
3 years ago
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