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Mathematics 7 Online
OpenStudy (anonymous):

help!!

OpenStudy (anonymous):

The dot plots below show the test scores of seventh- and eighth-grade students: src=" http://learn.flvs.net/webdav/assessment_images/educator_mjmath2_v14/mjmath2_practice_m8_g7_p.jpg " Based on visual inspection of the dot plots, which grade, if any, appears to have the higher mean score? Grade 7 Grade 8 Both groups show about the same mean score. No conclusion about mean score can be made from the data.

OpenStudy (phebe):

@Nnesha

OpenStudy (phebe):

i cant sorry

OpenStudy (phebe):

i dont even see your dotted line thing it said error

OpenStudy (phebe):

it say cookie error

OpenStudy (anonymous):

@markaskingalexandria1

OpenStudy (anonymous):

sorry cant help the page u added on ur question keeps saying error

OpenStudy (anonymous):

OMG!!

OpenStudy (anonymous):

wait do you have kik i cant take a pic and send it to you??

OpenStudy (anonymous):

*can

OpenStudy (anonymous):

nope i have hangout

OpenStudy (anonymous):

what is that?

OpenStudy (phebe):

i got both jus cant get on kik tho

OpenStudy (anonymous):

ok im downloading it right now

OpenStudy (anonymous):

ok whats your username

OpenStudy (anonymous):

@markaskingalexandria1

OpenStudy (anonymous):

for hangout

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

The price of some houses in a neighborhood are shown below: House Price A $120,000 B $130,000 C $140,000 D $150,000 E $1,110,000 Based on the data, should the mean or the median be used to make an inference about the price of the houses in the neighborhood? Median, because there is an outlier that affects the mean Mean, because there are no outliers that affect the mean Median, because it is in the center of the data Mean, because it is in the center of the data

OpenStudy (anonymous):

Median, because it is in the center of the data

OpenStudy (anonymous):

thank you!!

OpenStudy (anonymous):

The following table shows the data collected from a random sample of 100 middle school students on the number of hours they play outdoor games every week: Weekly Duration of Outdoor Games Time (in hours) 0–2 3–5 6–8 9–11 Number of Students 30 62 8 0 There are 1,200 students in the school. Based on the sample proportion, how many students in the school would be expected to play outdoor games for at least three hours every week? 744 840 960 1,104

OpenStudy (anonymous):

idk this one

OpenStudy (anonymous):

Jack wants to know how many families in his small neighborhood of 60 homes would help organize a neighborhood fund-raising party. He put all the addresses in a bag and drew a random sample of 30 addresses. He then asked those families if they would help organize the fund-raising party. He found that 12% of the families would help organize the party. He claims that 12% of the neighborhood families would be expected to help organize the party. Is this a valid inference? No, this is not a valid inference because he asked only 30 families No, this is not a valid inference because he did not take a random sample of the neighborhood Yes, this is a valid inference because he took a random sample of the neighborhood Yes, this is a valid inference because the 30 families speak for the whole neighborhood

OpenStudy (anonymous):

??

OpenStudy (anonymous):

Yes, this is a valid inference because he took a random sample of the neighborhood

OpenStudy (anonymous):

its still not letting me see it

OpenStudy (anonymous):

The following table shows the data collected from a random sample of 100 middle school students on the number of hours they play outdoor games every week: Weekly Duration of Outdoor Games Time (in hours) 0–2 3–5 6–8 9–11 Number of Students 30 62 8 0 There are 1,200 students in the school. Based on the sample proportion, how many students in the school would be expected to play outdoor games for at least three hours every week? 744 840 960 1,104

OpenStudy (anonymous):

@MTALHAHASSAN2

OpenStudy (anonymous):

@MTALHAHASSAN2 ??

OpenStudy (mtalhahassan2):

yes

OpenStudy (anonymous):

the answer..?

OpenStudy (mtalhahassan2):

wait a sec

OpenStudy (mtalhahassan2):

let me think

OpenStudy (anonymous):

ok

OpenStudy (mtalhahassan2):

I will say look at 3-5 hours, and the sample is from 100 students, and we have 62 students who play at least 3 - 5 hours a day. So that's 62%, just multiply 1200 by 62%.

OpenStudy (anonymous):

so whats the answer

OpenStudy (mtalhahassan2):

i am getting 744 but not sure if is it right or not

OpenStudy (anonymous):

ok, i got the same thing

OpenStudy (mtalhahassan2):

i will say plz conform from someone else

OpenStudy (mtalhahassan2):

by the which grade question is it??

OpenStudy (anonymous):

huh?

OpenStudy (mtalhahassan2):

@Michele_Laino

OpenStudy (michele_laino):

the experimental frequency is: 62/100=0.62, so the requested number is: (62/100)*1200=...?

OpenStudy (anonymous):

?

OpenStudy (michele_laino):

the proportion of student which play outdoor games is: 62/100, namely 62 students over 100 students in total

OpenStudy (anonymous):

so whats the answer

OpenStudy (mtalhahassan2):

744

OpenStudy (anonymous):

ok thanks!! mind helpig me with a few more?

OpenStudy (michele_laino):

now we have to write that proportion for a sample of 1200 students, so the requested number of students, which play outdorr games, for a sample of 1200 students, is: \[\Large \frac{{62}}{{100}} \cdot 1200 = ...?\]

OpenStudy (michele_laino):

outdoor*

OpenStudy (anonymous):

<img src=" http://learn.flvs.net/webdav/assessment_images/educator_mjmath2_v14/mjmath2_practice_m8_g8_c2.jpg " alt=" Which of the following inferences can be made using the dot plots? Group X has a higher mean than group Y. Group Y has a higher mean than group X. There is no overlap between the data. The range of each group is different.

OpenStudy (anonymous):

@Michele_Laino @MTALHAHASSAN2

OpenStudy (michele_laino):

I'm sorry I see only an error message

OpenStudy (mtalhahassan2):

same

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