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3 years ago

3/(x+2) - 2/(x-1) ≥ 0

Mathematics

1 answer:

GalinKa [24]3 years ago

4 0

The answer is −2<x<1 or x≥7

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3 years ago

a certain pharmaceutical company know that, on average 4% of a certain type of pill has an ingredient that is below the minimum

FinnZ [79.3K]

Using the binomial distribution, it is found that there is a 0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

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

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

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

In this problem:

The sample has 20 pills, hence $n = 20$ .

100 - 4 = 96% are acceptable, hence $p = 0.96$

The probability that <u>fewer that 5 in a sample of 20 pills</u> will be acceptable is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

In which

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

$P(X = 0) = C_{20,0}.(0.96)^{0}.(0.04)^{20} = 0$

$P(X = 1) = C_{20,1}.(0.96)^{1}.(0.04)^{19} = 0$

$P(X = 2) = C_{20,2}.(0.96)^{2}.(0.04)^{18} = 0$

$P(X = 3) = C_{20,3}.(0.96)^{3}.(0.04)^{17} = 0$

$P(X = 4) = C_{20,4}.(0.96)^{4}.(0.04)^{16} = 0$

0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

A similar problem is given at brainly.com/question/24863377

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2 years ago

A students work to solve 2 6/7 / 2/7 is shown below.

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A) the only mistake is changing the mixed number to an improper fraction .

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would be the high range

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What is the Anwser if a divide 33 into 2 groups so the ratio is 4 to 7

Naddika [18.5K]

You have to solve this algebraicly.
To do this set up the proportion and add x to each term.
You get 4x and 7x
Set this equal to 33
4x+7x=33
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Divide by 11
x=3
Now go back to 4x and 7x
Multiply them by x and get 12(4×3=12) and 21(7×3=21)
One group is 12 and the other is 21.

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3 years ago