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scZoUnD [109]
3 years ago
5

As part of a quality-control program, 4 batteries from a box of 17 is chosen at random for testing. In how many ways can this te

st batch be chosen?
Mathematics
1 answer:
belka [17]3 years ago
5 0

Answer:

This test batch can be chosen in 2380 ways

Step-by-step explanation:

The order in which the batteries are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

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)!}

In how many ways can this test batch be chosen?

4 batteries from a set of 17. So

C_{17,4} = \frac{17!}{4!(17-4)!} = 2380

This test batch can be chosen in 2380 ways

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Given n objects arranged in a row, a subset of these objects is called unfriendly if no two of its elements are consecutive. Sho
Vesna [10]

Complete question is;

Given n objects are arranged in a row. A subset of these objects is called unfriendly, if no two of its elements are consecutive. Show that the number of unfriendly subsets of a k-element set is ( n−k+1 )

( k )

Answer:

I've been able to prove that the number of unfriendly subsets of a k-element set is;

( n−k+1 )

( k )

Step-by-step explanation:

I've attached the proof that the number of unfriendly subsets of a k-element set is;

( n−k+1 )

( k )

6 0
4 years ago
In a population of adults the proportions of people in various marital status are: Single:35%, Married: 60%, Divorced: 5%. A per
vredina [299]

Answer:

7/20 0r 0.35

Step-by-step explanation:

Probability of been single = percentage of single adult / sum of percentages

sum of percentages = 35 + 60 + 5 = 100

35/100 = 7/20 or 0.35

8 0
3 years ago
Suppose that at Northside High, the number of hours per week that seniors spend on homework is approximately Normally distribute
Aliun [14]

Answer:

The mean of the sampling distribution of means for the 36 students is of 18.6 homework hours per week.

Step-by-step explanation:

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 question:

For the population, the mean is 18.6. So, by the Central Limit Theorem, the mean of the sampling distribution is also 18.6.

3 0
3 years ago
Two buses leave a station at the same time, traveling in opposite directions. The rate of the faster bus is 15 mph faster than t
ExtremeBDS [4]
A set of equation can be set up:
a=b+15
3(a+b)=345
where a is the faster and b is the slower bus.
substitute b+15 into the second equation, so 6b+45=345, and 6b=300, therefore b=50.
we can then figure out that a=65
5 0
3 years ago
Read 2 more answers
Complete the statement by filling in the blanks: When constructing a confidence interval, if the level of confidence increases,
Yuki888 [10]

Answer:

\hat p \pm z_{\alpha/2} SE_{\hat p}

And for this case the margin of error would be:

ME= z_{\alpha/2} SE_{\hat p}

If the level of confidence increase we can conclude that the value of z_{\alpha/2} would increase and the the confidence interval would be wider, since the margin of error increase.

c. Increase; wider

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Let's assume that we have a parameter of interest p and we want to estimate the value of p with \hat p and in general the confidence interval if the distribution of p is normal is given by:

\hat p \pm z_{\alpha/2} SE_{\hat p}

And for this case the margin of error would be:

ME= z_{\alpha/2} SE_{\hat p}

If the level of confidence increase we can conclude that the value of z_{\alpha/2} would increase and the the confidence interval would be wider, since the margin of error increase.

c. Increase; wider

8 0
3 years ago
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