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2 years ago

Hhhhhhhhhhhhhhhheeeeeeeeeeeeeeeelllllllllllllllllllllppppppppppppp

Mathematics

2 answers:

m_a_m_a [10]2 years ago

7 0

Literally whatever number is in the parenthesis, if it’s negative like -4 then you could down 4 lines of squares and put a dot. if it’s positive then go up

NikAS [45]2 years ago

6 0

What the person above me said :(()

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Find the distance between the two points rounding to the nearest tenth (if necessary).

vovangra [49]

Answer: 10.3

Step-by-step explanation:

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3 years ago

Consider a series system composed of 4 separate components where each component has a 30% chance of failing. Assume each compone

Marina86 [1]

Answer:

16.15% probability that exactly 3 of them would function

Step-by-step explanation:

Binomial probability 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.

Probability of each system working:

4 components, which means that $n = 4$

Each has a 30% probability of failing, so $p = 1 - 0.3 = 0.7$

For the system to work, all 4 components have to work. This is P(X = 4).

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

$P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401$

0.2401 probability of a system working.

If you have 7 of these systems, what is the probability that exactly 3 of them would function?

Now 7 systems, so $n = 7$

0.2401 probability of a system working.

We have to find P(X = 3).

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

$P(X = 3) = C_{7,3}.(0.2401)^{3}.(0.7599)^{4} = 0.1615$

16.15% probability that exactly 3 of them would function

5 0

3 years ago

Find AB If AC = 12 and BC = 35

Alisiya [41]

Answer:

Step-by-step explanation:

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2 years ago

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st

Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so $\mu = 0.5, \sigma = 0.05$ .