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poizon [28]
3 years ago
5

A container holds 0.7 liters of oil and vinegar 3/4 of that is vinegar how much oil is in the container

Mathematics
1 answer:
fgiga [73]3 years ago
3 0

Answer:

There are 21/40 L or 0.525 L of vinegar in the container.

Step-by-step explanation:

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How many significant figures are there In the number 10.76
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Will pick brainlyest help me plz 15 pts
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1.50c + 2.50a = 1100
c + a = 500

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3 years ago
Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose
Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

In which

x is the number of sucesses

&#10;e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Poisson mean:

\mu = 0.2

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

P(X \geq 2) = 1 - P(X < 2)

In which

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

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so p = 0.181

We want to find P(X = 0).

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

P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671

0.671 = 67.1% probability that neither contains a missing pulse

8 0
3 years ago
What is the easiest way to get brainiest? I have 1000 points and I'm only missing the brainiest points. I am still rank Ambitiou
Alexus [3.1K]

Answer:

Very nice to see someone ambitious. You need to answer thouroghly in order to get brainliest. Explain every step, as tiring as it may seem.

Have a good day!

5 0
3 years ago
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