The post Li designed a survey to determine how comfortable students at her middle school are with fractions. There are 600 students in her middle school. The students are equally distributed among grade levels. She selected a sample of 28 students in her first-period class. Which best explains Li’s sample? appeared first on EduHawks.com.

]]>The correct answer to this is:

**Nonrandom and biased**

The method of Li’s sampling has a

tendency that some elements or portion in the population has no chance of

selection. Further, it was not stated which portion of the population she

obtained the sample thus we cannot accurately determine the probability of

selection. This kind of sampling results in a bias. A biased sampling is a type

of nonrandom sampling.

The post Li designed a survey to determine how comfortable students at her middle school are with fractions. There are 600 students in her middle school. The students are equally distributed among grade levels. She selected a sample of 28 students in her first-period class. Which best explains Li’s sample? appeared first on EduHawks.com.

]]>The post A. about 60% of florida residents believe that florida is a nice place to live. suppose that six randomly selected florida residents are interviewed. what is the probability that at least one resident does not think that florida is a nice place to live? appeared first on EduHawks.com.

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At least 6, means: either 1 and 2 and 3 and 4 and 5 and 6

Probability(that Florida is a nice place to live) = 60% = 0.6

If at least 6 randomly selected like Florida:

P(at lest 6 LIKE Florida) = (0.6)⁶ = 0.04665

P(at least 6 DON’T LIKE Florida) = 1-0.04665 = 0.9533

The post A. about 60% of florida residents believe that florida is a nice place to live. suppose that six randomly selected florida residents are interviewed. what is the probability that at least one resident does not think that florida is a nice place to live? appeared first on EduHawks.com.

]]>The post Samples of size n = 90 are randomly selected from the population of numbers (0 through 9) produced by a random-number generator, and the variance is found for each sample. What is the distribution of the sample variances? appeared first on EduHawks.com.

]]>**Answer:**

chi square distribution with df = n-1

**Step-by-step explanation:**

Given that sample of size n=90 are randomly selected from the population of numbers (0 through 9) produced by a random-number generator

Since sample size is large and randomness is followed we can assume that the variable follows a normal distribution.

Hence the sample variance would follow a chi square distribution with degree of freedom =

This is because we have is standard normal hence square will be a chisquare variate. When we sum n variates we get chi square distribution with df n-1

The post Samples of size n = 90 are randomly selected from the population of numbers (0 through 9) produced by a random-number generator, and the variance is found for each sample. What is the distribution of the sample variances? appeared first on EduHawks.com.

]]>The post What feature of excel allows you to automatically calculate common formulas with selected data appeared first on EduHawks.com.

]]>What feature of excel allows you to automatically calculate common formulas with selected data

The post What feature of excel allows you to automatically calculate common formulas with selected data appeared first on EduHawks.com.

]]>The post Read the sentence and answer the question. Nine more pilots were selected in 1962 fourteen were chosen in 1963. Which correctly shows where a semicolon is needed in the sentence? Nine more pilots were selected; in 1962 fourteen were chosen in 1963. Nine more pilots were selected in 1962 fourteen were chosen; in 1963. Nine more pilots were selected in 1962; fourteen were chosen in 1963. Nine more pilots; were selected in 1962 fourteen were chosen in 1963. Description appeared first on EduHawks.com.

]]>Read the sentence and answer the question. Nine more pilots were selected in 1962 fourteen were chosen in 1963. Which correctly shows where a semicolon is needed in the sentence? Nine more pilots were selected; in 1962 fourteen were chosen in 1963. Nine more pilots were selected in 1962 fourteen were chosen; in 1963. Nine more pilots were selected in 1962; fourteen were chosen in 1963. Nine more pilots; were selected in 1962 fourteen were chosen in 1963. Description

The post Read the sentence and answer the question. Nine more pilots were selected in 1962 fourteen were chosen in 1963. Which correctly shows where a semicolon is needed in the sentence? Nine more pilots were selected; in 1962 fourteen were chosen in 1963. Nine more pilots were selected in 1962 fourteen were chosen; in 1963. Nine more pilots were selected in 1962; fourteen were chosen in 1963. Nine more pilots; were selected in 1962 fourteen were chosen in 1963. Description appeared first on EduHawks.com.

]]>The post Consider a political discussion group consisting of 9 Democrats, 7 Republicans, and 3 Independents. Suppose that two group members are randomly selected, in succession, to attend a political convention. Find the probability of selecting no Democrats appeared first on EduHawks.com.

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Well, first you would add up how many objects you have to select from

9 democrats + 7 Republicans + 3 Independents = 19 total people

so then you would put how many people you can have over that number

so since we don’t want any Democrats it would be 10/19 (.52)

and that tells you the chances of not picking one the first time

then, since that person cannot be selected again, we remove them from the pool, making it only 18 people to pick from and 9 people you want picked so once again you put 9/18 (.50), then you take those, (in decimal form) and multiply them (.26) then convert it to a fraction by moving the decimal to the right two and have 26% chance of no Democrats

The post Consider a political discussion group consisting of 9 Democrats, 7 Republicans, and 3 Independents. Suppose that two group members are randomly selected, in succession, to attend a political convention. Find the probability of selecting no Democrats appeared first on EduHawks.com.

]]>The post The body temperatures of adults are normally distributed with a mean of 98.6° f and a standard deviation of 0.60° f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° f. 0.0228 0.8188 0.9360 0.9772 appeared first on EduHawks.com.

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Here the population standard deviation is 0.60 degree F. If a sample of 36 adults is randomly selected, that results in a sample standard deviation of 0.60 degree F divided by the square root of 36: 0.10 degree F.

The probability in question is the area under the standard normal probability distribution between 98.4 degree F and infinity, and intuitively you can detect that this will be more than 0.5 (corresponding to 50%).

Convert 98.4 degrees F to a z-score, using the sample standard deviation (0.10 degree F). That z score is

98.4-98.6

z = ————– = -0.20/0.10 = -2

0.10

We need to determine the area under the standard normal curve to the right of z=-2. Use a table of z-scores to do this, or use your calculator’s built-in probability functions. My result is 98.21% (corresponding to an area of 0.9821).

With my calculator I can find this probability using the following command:

normalcdf(-2,100000,0.10).

The post The body temperatures of adults are normally distributed with a mean of 98.6° f and a standard deviation of 0.60° f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° f. 0.0228 0.8188 0.9360 0.9772 appeared first on EduHawks.com.

]]>The post When campaign workers are canvassing neighborhoods, one of their goals is to identify voters who will support their candidate. sell campaign buttons door to door. determine the income level of a community. find out which candidate voters have selected? appeared first on EduHawks.com.

]]>**Answer:**

The correct answer here is **to identify voters who will support their candidate**.

**Explanation:**

Door-to-door campaigning is a process and a method that has been used by a lot of different organizations because it allows for a more personalized and individualized contact so that messages can be conveyed in a more personal manner. But probably, one of the sectors that has used this method more is the political sector, especially during political campaigns. This is because it allows for candidates to, through their representatives, have closer contact with their potential voters, it allows for messages to be delivered directly to individuals, instead of in mass, which can also help convince those who are undecided, and most importantly, this method allows candidates to rally their voters, seem closer to them and generate a voter group who will support the candidate.

The post When campaign workers are canvassing neighborhoods, one of their goals is to identify voters who will support their candidate. sell campaign buttons door to door. determine the income level of a community. find out which candidate voters have selected? appeared first on EduHawks.com.

]]>The post A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. if 2 dogs are selected at random, what is the probability that both selected dogs are not littermates? appeared first on EduHawks.com.

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You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.

So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:

C (m,n) = m! / (n! * (m -n)! )

=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =

= **47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.**

The post A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. if 2 dogs are selected at random, what is the probability that both selected dogs are not littermates? appeared first on EduHawks.com.

]]>The post A math teacher wants to find out the average number of hours his students spend working on their math homework for his class each week. Which group would best represent a sample of the population? the first 30 students who enter the school building one morning 30 students selected from the lunchroom 30 students selected from his class rosters 30 students who stay after school for football practice appeared first on EduHawks.com.

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You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.

So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:

C (m,n) = m! / (n! * (m -n)! )

=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =

= **47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.**

The post A math teacher wants to find out the average number of hours his students spend working on their math homework for his class each week. Which group would best represent a sample of the population? the first 30 students who enter the school building one morning 30 students selected from the lunchroom 30 students selected from his class rosters 30 students who stay after school for football practice appeared first on EduHawks.com.

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