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Mathematics 8 Online
OpenStudy (muscrat123):

@heretohelpalways

OpenStudy (anonymous):

yes?

OpenStudy (muscrat123):

Maggie has 7 tiles with pictures of plants and 2 tiles with pictures of animals. Maggie keeps all the tiles on a mat with the pictures hidden and mixes them up. She then turns one tile face up and finds the picture of a plant on it. She removes this tile from the mat and turns over another tile without looking. What is the probability that the second tile that Maggie turns over has a plant on it? 22.2% 28.6% 65.0% 75.0% i think it may be A

OpenStudy (anonymous):

the new probability would be 6/8

OpenStudy (muscrat123):

so D?

OpenStudy (anonymous):

which is equal to....

OpenStudy (anonymous):

yup D

OpenStudy (muscrat123):

Ben has a 40% chance of being selected as the president of the school sports club and a 70% chance of being elected editor of the school magazine. What is the probability that Ben will be selected as the president of the sports club and will also be elected editor of the school magazine? 0.3% 1.1% 14% 28%

OpenStudy (muscrat123):

i dont know how to solve problems like this

OpenStudy (anonymous):

do 4 over 10 * 7 over 10

OpenStudy (anonymous):

\[\frac{ 4 }{ 10 }*\frac{ 7 }{ 10 }\]

OpenStudy (anonymous):

28/100

OpenStudy (muscrat123):

28%

OpenStudy (anonymous):

yup

OpenStudy (muscrat123):

Max tossed a fair coin 3 times. What is the probability of getting heads in the first two trials and tails in the last trial? 4 over 8 3 over 8 2 over 8 1 over 8

OpenStudy (anonymous):

do \[\frac{ 1 }{ 2 }*\frac{ 1 }{ 2 }*\frac{ 1 }{ 2 }\]

OpenStudy (anonymous):

I think

OpenStudy (anonymous):

I am really not sure about this one

OpenStudy (anonymous):

1/8

OpenStudy (muscrat123):

but 13% isnt an option

OpenStudy (anonymous):

1/8 is an option...

OpenStudy (muscrat123):

oh we are talking about the most recent question. i thought we were talking about the plant question

OpenStudy (anonymous):

|dw:1438870916332:dw|

OpenStudy (muscrat123):

At a game booth, a student gets a box of candy as the prize for winning a game. The boxes come in four colors: white, red, green, and blue. There are 10 boxes of each color. All the boxes are equally likely to be given away as prizes. Which expression shows the probability of the first winner receiving a white box and the second winner also receiving a box of the same color? 10 over 40 multiplied by 9 over 39 10 over 40 multiplied by 10 over 39 10 over 40 plus 9 over 39 10 over 40 plus 10 over 39

OpenStudy (anonymous):

you must NOT add these fractions, so that should eliminate 2 answer

OpenStudy (anonymous):

and a white box was taken out leaving the probability of a white box to be chosen 9/39

OpenStudy (anonymous):

so it would be A

OpenStudy (anonymous):

anymore questions?

OpenStudy (muscrat123):

sorry it took a while to respond - had to do an interview for flvs A box has 5 beads of the same size, but all are different colors. Tina draws a bead randomly from the box, notes its color, and then puts the bead back in the box. She repeats this 3 times. What is the probability that Tina would pick a red bead on the first draw, then a green bead, and finally a red bead again? 1 over 625 1 over 180 1 over 150 1 over 125 @heretohelpalways

OpenStudy (anonymous):

that's okay

OpenStudy (anonymous):

\frac{ 1 }{ 5 }*\frac{ 1 }{ 5 }*\frac{ 1 }{ 5 }

OpenStudy (anonymous):

do 1/5*1/5*1/5

OpenStudy (anonymous):

to get 1/125

OpenStudy (anonymous):

was the interview a DBA?

OpenStudy (muscrat123):

b4 the interview yes

OpenStudy (anonymous):

lol

OpenStudy (muscrat123):

i had to do an actual interview though

OpenStudy (muscrat123):

could u help with some more?

OpenStudy (anonymous):

sure

OpenStudy (muscrat123):

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):

can you make the table a bit cleaner?

OpenStudy (anonymous):

do 62+8+0

OpenStudy (anonymous):

to get 70%

OpenStudy (anonymous):

find 70% of 1,200

OpenStudy (anonymous):

to get the answer: 840

OpenStudy (muscrat123):

so the final answer is 840?

OpenStudy (anonymous):

yup

OpenStudy (muscrat123):

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):

it is valid since he took a random sample

OpenStudy (muscrat123):

The price of some jewelry sets in a store are shown below: Store Price A $110,000 B $100,000 C $1,110,000 D $130,000 E $120,000 Based on the data, should the mean or the median be used to make an inference about the price of the jewelry sets in the store? Mean, because it is in the center of the data Median, because it is in the center of the data Median, because there is an outlier that affects the mean Mean, because there are no outliers that affect the mean @heretohelpalways

OpenStudy (anonymous):

median, since outliers don't affect medians so C

OpenStudy (muscrat123):

Noah wants to predict who will be selected as the sixth-grade treasurer for student council. Which of the following groups should Noah survey to make his prediction? The first 100 students from an alphabetical list of the entire school The first 100 students from an alphabetical list of sixth graders One hundred random students from the entire school One hundred random students from the sixth grade D correct @heretohelpalways ?

OpenStudy (muscrat123):

and also: Four students did a survey to find the pizza topping sixth-grade students prefer. The table below shows the method each student used to conduct the survey: Student Method Troy Asked 100 students at random from his seventh-grade class what their favorite pizza topping is Dacia Asked 100 eighth-grade students at random what their favorite pizza topping is Julia Asked 100 sixth-grade students at random what their favorite pizza topping is Tanner Asked 100 third-grade students at random what their favorite pizza topping is Which student's survey is most likely not biased? Troy Dacia Julia Tanner

OpenStudy (anonymous):

D is correcct

OpenStudy (anonymous):

and JUlia is not biased because she is the only one who asked sixth graders

OpenStudy (anonymous):

oops Julia

OpenStudy (muscrat123):

The following table shows the data collected from a random sample of 100 middle school students on the number of hours they play indoor games every week: Weekly Duration of Indoor Games Time (in hours) 0–3 4–7 8–11 12–15 Number of Students 4 42 52 2 There are 1,300 students in the school. Based on the sample proportion, how many students in the school would be expected to play indoor games for at least four hours every week? 546 598 702 1,248 \ @heretohelpalways

OpenStudy (anonymous):

ooops

OpenStudy (anonymous):

96% of 1,300

OpenStudy (anonymous):

1248

OpenStudy (muscrat123):

The dot plot below shows the amount of time two random groups of students took to brush their teeth: Based on visual inspection of the dot plots, which of the following groups, if any, shows a greater average time required to brush their teeth? Group R Group S Both groups show about the same average time. No conclusion about average time can be made from the data. @heretohelpalways

OpenStudy (muscrat123):

group s?

OpenStudy (anonymous):

I am not really sure...

OpenStudy (muscrat123):

The price of some jewelry sets in a store are shown below: Store Price A $110,000 B $100,000 C $1,110,000 D $130,000 E $120,000 Based on the data, should the mean or the median be used to make an inference about the price of the jewelry sets in the store? Mean, because it is in the center of the data Median, because it is in the center of the data Median, because there is an outlier that affects the mean Mean, because there are no outliers that affect the mean C?

OpenStudy (anonymous):

try calculating the mean

OpenStudy (anonymous):

of each

OpenStudy (muscrat123):

The price of some jewelry sets in a store are shown below: Store Price A $110,000 B $100,000 C $1,110,000 D $130,000 E $120,000 Based on the data, should the mean or the median be used to make an inference about the price of the jewelry sets in the store? Mean, because it is in the center of the data Median, because it is in the center of the data Median, because there is an outlier that affects the mean Mean, because there are no outliers that affect the mean C? and document.rtf is a different question

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