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3 years ago

For the set C={3,5}, determine n(C) n(C)=

Mathematics

2 answers:

Otrada [13]3 years ago

3 0

Answer: 2

The notation n(C) translates to "the number of items in set C". So all that you do is count the number of values in the set {3,5} which happens to be two items.

Some books use absolute value notation to indicate the number of items. So some books may write |C| instead of n(C). This concept is known as set cardinality (or the cardinal value of a set).

Nonamiya [84]3 years ago

3 0

Answer:

n(C) = 2

Step-by-step explanation:

A set is given as C = {3, 5}. That means set C is having tow elements.

n(C) is referring the number of elements of the set C.

n(C) is also known as cardinality of the set.

In the given set number of elements are two so cardinality of the set n(C) is two.

Therefore, n(C) = 2 will be the answer.

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At a cell phone assembly plant, 77% of the cell phone keypads pass inspection. A random sample of 111 keypads is analyzed. Find

Serggg [28]

Answer:

The probability that the proportion of passed keypads is between 0.72 and 0.80 is 0.6677.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

Let p = the proportion of keypads that pass inspection at a cell phone assembly plant.

The probability that a randomly selected cell phone keypad passes the inspection is, p = 0.77.

A random sample of n = 111 keypads is analyzed.

Then the sampling distribution of $\hat p$ is:

$\hat p\sim N(p,\ \sqrt{\frac{p(1-p)}{n}})\\\Rightarrow \hat p\sim N (0.77, 0.04)$

Compute the probability that the proportion of passed keypads is between 0.72 and 0.80 as follows:

$P(0.72$

$=P(-1.25$

Thus, the probability that the proportion of passed keypads is between 0.72 and 0.80 is 0.6677.

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