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3 years ago

For the set D = (4,6,8,10), determine n(D)

Mathematics

1 answer:

pentagon [3]3 years ago

8 0

n(D) = 4

Step-by-step explanation:

Given Data:

Set D = {4, 6, 8, 10}

The number of elements in a set is called the cardinal elements (or) cardinality of the sets. This is denoted by n(A).

Here it is denoted by n(D).

The number of elements in the set D is 4.

The cardinality of the set is 4.

Hence, n(D) = 4.

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HELP PLS I NEED THIS RIGHT NOW

Ne4ueva [31]

Answer:

period = 2

Step-by-step explanation:

The period is the measure of 1 complete cycle of the wave

The wave enters at x = 0 , y = - 0.5

After 1 cycle the wave is at x = 2, y = - 0.5

period = 2 - 0 = 2

6 0

3 years ago

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

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

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago

How much smaller is the sum of the first 1000 natural numbers than the sum of the first 1001 natural numbers?

aleksandr82 [10.1K]

ANSWER

1001

EXPLANATION

The sum of the first n - natural numbers is

$S_n= \frac{n}{2} (2a + d(n - 1))$