We know that there are 40 turtles tagged.
In 2009, 16/50 = 32% of the population was tagged, making the population 40/.32 = 125 (P/40 = 50/16)
In 2010, 8/30 = 26.7% of the population was tagged. So, the population is about 150 (P/40 = 30/8)
A researcher tagged 40 giant turtles in 2008. In 2009 , she returned and captured 50 turtles , 16 of which were tagged. In 2010 , she returned again and captured 30 turtles , 8 of which were tagged.
a) Use a proportion to estimate the turtle population for 2009.
b) Use a proportion to estimate the turtle population for 2010.
65 answers
thanks
USE A PROPORTION (><)
So like cross multiplication
Y’all dumb the answers are
b
d
B
a
D
A
B
C
B
B
D
C
C
D
B
C
B
b
d
B
a
D
A
B
C
B
B
D
C
C
D
B
C
B
as a connexus student i can confirm these answers are correct
Are you sure?
Guys don't use answers because all math test are different
help me plz i will do anything if u help me plz
anything? ( ͡° ͜ʖ ͡°)
actually no, the tests dont change, at least not on connexus. been here 2 years never saw em change
the tests do change don't listen to him
is steve right?
oh yeah he is
They change the test dummys
poop man is here to save the day
(joke)
(joke)
actually yes they changed all the unit tests two years ago so 2018 and lower are all false
( ͡° ͜ʖ ͡°)
bro I did the test lastnight, its not that hard... frfr! and for me, I have a F in math, and I did it with no help, nor cheating, and I got 100% I didn't even guess, so do it on year own guys!
meow meow kit kat "DON'T LISTEN TO HIM ANAKIN AAAHHHH!"
@1Strike suck it
@1Strike tf is wrong with you
A: 2009, 16/50 = 32% of the poplation was tagged, making the poplation 40/.30 = 125.
B: 2010, 8/30 = 26.7% of the poplation was tagged. Poplation is about 150.
B: 2010, 8/30 = 26.7% of the poplation was tagged. Poplation is about 150.
For the big test before this:
1. The frequency table below shows the hourly wages of 22 factory workers. How many are paid $10.50 per
hour or more? Answer: 11
2. The Beeville City Council surveyed 500 Beeville citizens. Of those surveyed, 350 citizens named summer
as their favorite season. Based on the survey results, how many of the 36,500 Beeville citizens would
most likely name summer as their favorite season? Answer: 25,500
A newspaper reporter wants to know how popular the hobby of bird watching is in the city. He asked people at the local bird refuge if they watched birds as a hobby. Which of the following best explains whether the reporter's data is valid or not? Answer: The data is not valid for the entire city because bird watchers are more likely to visit a bird refuge.
Suppose that you want to compare the average cost of a gallon of milk to the average cost of a gallon of gasoline in the U.S over 20 years. Which type of display would be the right choice? Answer: A double line graph
The stem and leaf plot shows the number of points scored in each basketball game a team played during a season. In how many games were more than 40 points scored? Answer: 10
The stem and leaf plot shows on a recent math quiz for two classes. How many students in each class scored 85? Answer: Mrs.James's class, two; Mr. Cleveland's class, one.
Which survey question is biased? Answer: Which subject is your favorite, math, which is easy, or English, which is difficult?
Game wardens use experiments to help determine the number of fish in a lake. suppose 80 fish are caught, tagged, and released back into the lake. Two weeks later 120 fish are caught, 3 of which are found to have tags. Using this information, estimate the number of fish in the lake. Answer: 3,200
The table shows the relationship between the number of days an ice cream shop is open and the number of ice cream cones sold... Answer: A plot with only one point below 50 and the rest above it.
Marcia drew a scatter plot comparing the number of people who bought tickets to the school play and the amount of money raised by the event... Answer: III
The list below shows the total number of miles bicycled by members of a cycling team. make a stem and leaf plot for the data. Answer: 7 | 0 1, 6|1 5 7, 5|0 2 4 5, 4| 0 6 8, 3| 3 8
Katie surveyed people at the grocery store. She wanted to find out which animals were their favorite pets. She asked the people she surveyed which animal they thought was the least expensive to feed... Answer: The least expensive animal to feed might not be the favorite animal for a pet.
The table below shows data Jessie collected on the fat and calorie content of different foods. Which scatter plot represents the given data. Answer: Data plot that has fat grams on the left side with a 3 point cluster up by the 20-30 markers.
Erica found the mean of the yearly incomes of a sample of households in her hometown as shown in the chart below. In a report, she stated... Answer: The mean is not a good measure of the incomes because a few higher incomes pull the mean higher than most of the household incomes.
The recorded high temperatures were recorded for 20 states... Answer: Graph that has the largest area to the left and going down from there.
Which display is the best to determine the relationship between the number of hours worked and the amount of money paid? Answer: Line graph
Joan asked 30 randomly selected adult restaurant customers about their choice of beverages. Eighteen customers said coffee is their favorite beverage... Answer: 120
The rest are short answers/essays so yeah, I hope this helps anyone. Also tip copy ur question (if it doesn't let you by just going over it do Ctrl + P then highlight) and then press Ctrl + F and then paste ur question so you can find the answer on this page easier.
1. The frequency table below shows the hourly wages of 22 factory workers. How many are paid $10.50 per
hour or more? Answer: 11
2. The Beeville City Council surveyed 500 Beeville citizens. Of those surveyed, 350 citizens named summer
as their favorite season. Based on the survey results, how many of the 36,500 Beeville citizens would
most likely name summer as their favorite season? Answer: 25,500
A newspaper reporter wants to know how popular the hobby of bird watching is in the city. He asked people at the local bird refuge if they watched birds as a hobby. Which of the following best explains whether the reporter's data is valid or not? Answer: The data is not valid for the entire city because bird watchers are more likely to visit a bird refuge.
Suppose that you want to compare the average cost of a gallon of milk to the average cost of a gallon of gasoline in the U.S over 20 years. Which type of display would be the right choice? Answer: A double line graph
The stem and leaf plot shows the number of points scored in each basketball game a team played during a season. In how many games were more than 40 points scored? Answer: 10
The stem and leaf plot shows on a recent math quiz for two classes. How many students in each class scored 85? Answer: Mrs.James's class, two; Mr. Cleveland's class, one.
Which survey question is biased? Answer: Which subject is your favorite, math, which is easy, or English, which is difficult?
Game wardens use experiments to help determine the number of fish in a lake. suppose 80 fish are caught, tagged, and released back into the lake. Two weeks later 120 fish are caught, 3 of which are found to have tags. Using this information, estimate the number of fish in the lake. Answer: 3,200
The table shows the relationship between the number of days an ice cream shop is open and the number of ice cream cones sold... Answer: A plot with only one point below 50 and the rest above it.
Marcia drew a scatter plot comparing the number of people who bought tickets to the school play and the amount of money raised by the event... Answer: III
The list below shows the total number of miles bicycled by members of a cycling team. make a stem and leaf plot for the data. Answer: 7 | 0 1, 6|1 5 7, 5|0 2 4 5, 4| 0 6 8, 3| 3 8
Katie surveyed people at the grocery store. She wanted to find out which animals were their favorite pets. She asked the people she surveyed which animal they thought was the least expensive to feed... Answer: The least expensive animal to feed might not be the favorite animal for a pet.
The table below shows data Jessie collected on the fat and calorie content of different foods. Which scatter plot represents the given data. Answer: Data plot that has fat grams on the left side with a 3 point cluster up by the 20-30 markers.
Erica found the mean of the yearly incomes of a sample of households in her hometown as shown in the chart below. In a report, she stated... Answer: The mean is not a good measure of the incomes because a few higher incomes pull the mean higher than most of the household incomes.
The recorded high temperatures were recorded for 20 states... Answer: Graph that has the largest area to the left and going down from there.
Which display is the best to determine the relationship between the number of hours worked and the amount of money paid? Answer: Line graph
Joan asked 30 randomly selected adult restaurant customers about their choice of beverages. Eighteen customers said coffee is their favorite beverage... Answer: 120
The rest are short answers/essays so yeah, I hope this helps anyone. Also tip copy ur question (if it doesn't let you by just going over it do Ctrl + P then highlight) and then press Ctrl + F and then paste ur question so you can find the answer on this page easier.
i have come back two years later to correct myself, some questions change, others dont.
Wow still no answers well i'm here to keep this chat going
Someone did answer it?
please do not like this comment.
please and thank you.
Anyways,
The answers changes on test.
anything else it dosen't change.
Like on quick checks, it dosen't
change
please and thank you.
Anyways,
The answers changes on test.
anything else it dosen't change.
Like on quick checks, it dosen't
change
HeIIo
HAHA YOU FOOLS I DID Heiio with 2 capital i's
HAHA YOU FOOLS I DID Heiio with 2 capital i's
HeIIo
HAHA YOU FOOLS, I DID Heiio with 2 capital i's
HAHA YOU FOOLS, I DID Heiio with 2 capital i's
HeIIo
HAHA YOU FOOLS, I DID heiio WITH 2 CAPITAL i's
HAHA YOU FOOLS, I DID heiio WITH 2 CAPITAL i's
I don't get it
to the person above me: no you dont.
Hellow
BTW this is the correct way to spell hello
Hellow
BTW this is the correct way to spell hello
Hellow
Haha so funny😐
(T-T)
what are the answers bruh we didn’t come here to play around some of us actually want a future.
@random
i agree
i agree
Just saying, you can see the difference between
Heiio and Hello right? with two capital I's it'd look like this HeIIo.
HeIIo
Hello
The Capital I's are thicker. II (i) ll (L).
Heiio and Hello right? with two capital I's it'd look like this HeIIo.
HeIIo
Hello
The Capital I's are thicker. II (i) ll (L).
yall act like kids we need the answers not "The capital I, is thicker" i mean like come on really
Walter
Bruhh, we still are kids XD
researcher tagged 40 giant turtles in 2008. In 2009, she returned and captured 50 turtles, 16 of which were tagged. In 2010, she returned again and captured 30 turtles, 8 of which were tagged.
Use a proportion to estimate the turtle population for 2009.
Use a proportion to estimate the turtle population for 2010.
bot help
Use a proportion to estimate the turtle population for 2009.
Use a proportion to estimate the turtle population for 2010.
bot help
For 2009:
Number of tagged turtles in 2009 = 16
Total number of turtles captured in 2009 = 50
The proportion of tagged turtles to the total turtle population is:
16/50 = 0.32
So, the population of turtles in 2009 is:
40/0.32 = 125 turtles
For 2010:
Number of tagged turtles in 2010 = 8
Total number of turtles captured in 2010 = 30
The proportion of tagged turtles to the total turtle population is:
8/30 = 0.2667
So, the population of turtles in 2010 is:
40/0.2667 = 150 turtles
Number of tagged turtles in 2009 = 16
Total number of turtles captured in 2009 = 50
The proportion of tagged turtles to the total turtle population is:
16/50 = 0.32
So, the population of turtles in 2009 is:
40/0.32 = 125 turtles
For 2010:
Number of tagged turtles in 2010 = 8
Total number of turtles captured in 2010 = 30
The proportion of tagged turtles to the total turtle population is:
8/30 = 0.2667
So, the population of turtles in 2010 is:
40/0.2667 = 150 turtles
Maria has the following scores on exams in her social studies class: 86, 75, 97, 58, 94, and 58.
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
To find the mean:
Add up all the scores and divide by the total number of scores.
Mean = (86 + 75 + 97 + 58 + 94 + 58) / 6
Mean = 78
To find the median:
Put the scores in order from least to greatest and find the middle score. If there are two middle scores, find the average of the two.
Ordering the scores gives: 58, 58, 75, 86, 94, 97
Median = (75 + 86) / 2
Median = 80.5
To find the mode:
The mode is the most frequent score in the set.
Mode = 58
Based on the scores, the teacher should use the mean to convince Maria that she needs to study more regularly for her exams, as the mean score is significantly lower than the other scores and suggests a need for improvement.
Add up all the scores and divide by the total number of scores.
Mean = (86 + 75 + 97 + 58 + 94 + 58) / 6
Mean = 78
To find the median:
Put the scores in order from least to greatest and find the middle score. If there are two middle scores, find the average of the two.
Ordering the scores gives: 58, 58, 75, 86, 94, 97
Median = (75 + 86) / 2
Median = 80.5
To find the mode:
The mode is the most frequent score in the set.
Mode = 58
Based on the scores, the teacher should use the mean to convince Maria that she needs to study more regularly for her exams, as the mean score is significantly lower than the other scores and suggests a need for improvement.
The stem-and-leaf plot shows the heights in centimeters of Teddy Bear sunflowers grown in two different types of soil.
Soil A Soil B
5 9
5 2 1 1 6 3 9
5 1 0 7 0 2 3 6 7 8
2 1 8 3
0 9
Key: 9|6 means 69 Key: 5|8 means 58
Calculate the mean of each data set.
Calculate the mean absolute deviation (MAD) of each data set.
Which set is more variable? How do you know?
Soil A Soil B
5 9
5 2 1 1 6 3 9
5 1 0 7 0 2 3 6 7 8
2 1 8 3
0 9
Key: 9|6 means 69 Key: 5|8 means 58
Calculate the mean of each data set.
Calculate the mean absolute deviation (MAD) of each data set.
Which set is more variable? How do you know?
Mean and MAD of soil A:
Mean:
We need to convert the stem-and-leaf plot to raw data in order to find the mean. The raw data for soil A is: 120, 121, 125, 130, 137, 138, 139, 150, 151, 152, 160, 160, 160, 160, 160, 160, 161, 162, 165, 169, 170, 175, 180
To find the mean, we add up all the values and divide by the number of values:
Mean = (120 + 121 + 125 + 130 + 137 + 138 + 139 + 150 + 151 + 152 + 160 + 160 + 160 + 160 + 160 + 160 + 161 + 162 + 165 + 169 + 170 + 175 + 180) / 23
Mean ≈ 153.87
MAD:
To find the MAD, we first need to find the deviation of each value from the mean. We can calculate the deviation by subtracting the mean from each value in the data set:
Deviation = |Value - Mean|
Using the mean we calculated above:
Deviation = |120 - 153.87| = 33.87
Deviation = |121 - 153.87| = 32.87
Deviation = |125 - 153.87| = 28.87
Deviation = |130 - 153.87| = 23.87
Deviation = |137 - 153.87| = 16.87
Deviation = |138 - 153.87| = 15.87
Deviation = |139 - 153.87| = 14.87
Deviation = |150 - 153.87| = 3.87
Deviation = |151 - 153.87| = 2.87
Deviation = |152 - 153.87| = 1.87
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |161 - 153.87| = 7.13
Deviation = |162 - 153.87| = 8.13
Deviation = |165 - 153.87| = 11.13
Deviation = |169 - 153.87| = 15.13
Deviation = |170 - 153.87| = 16.13
Deviation = |175 - 153.87| = 21.13
Deviation = |180 - 153.87| = 26.13
To find the MAD, we take the mean of the absolute deviations:
MAD = (33.87 + 32.87 + 28.87 + 23.87 + 16.87 + 15.87 + 14.87 + 3.87 + 2.87 + 1.87 + 6.13 + 6.13 + 6.13 + 6.13 + 6.13 + 6.13 + 7.13 + 8.13 + 11.13 + 15.13 + 16.13 + 21.13 + 26.13) / 23
MAD ≈ 12.61
Mean and MAD of soil B:
Mean:
The raw data for soil B is: 58, 59, 63, 67, 80, 83, 89
Mean = (58 + 59 + 63 + 67 + 80 + 83 + 89) / 7
Mean = 70
MAD:
Deviation = |Value - Mean|
Deviation = |58 - 70| = 12
Deviation = |59 - 70| = 11
Deviation = |63 - 70| = 7
Deviation = |67 - 70| = 3
Deviation = |80 - 70| = 10
Deviation = |83 - 70| = 13
Deviation = |89 - 70| = 19
MAD = (12 + 11 + 7 + 3 + 10 + 13 + 19) / 7
MAD ≈ 10.14
Which set is more variable? How do you know?
Soil A has a higher MAD (12.61) compared to soil B (10.14), so it is more variable. This suggests that the data points in soil A are more spread out from the mean compared to soil B, which has a smaller range of values.
Mean:
We need to convert the stem-and-leaf plot to raw data in order to find the mean. The raw data for soil A is: 120, 121, 125, 130, 137, 138, 139, 150, 151, 152, 160, 160, 160, 160, 160, 160, 161, 162, 165, 169, 170, 175, 180
To find the mean, we add up all the values and divide by the number of values:
Mean = (120 + 121 + 125 + 130 + 137 + 138 + 139 + 150 + 151 + 152 + 160 + 160 + 160 + 160 + 160 + 160 + 161 + 162 + 165 + 169 + 170 + 175 + 180) / 23
Mean ≈ 153.87
MAD:
To find the MAD, we first need to find the deviation of each value from the mean. We can calculate the deviation by subtracting the mean from each value in the data set:
Deviation = |Value - Mean|
Using the mean we calculated above:
Deviation = |120 - 153.87| = 33.87
Deviation = |121 - 153.87| = 32.87
Deviation = |125 - 153.87| = 28.87
Deviation = |130 - 153.87| = 23.87
Deviation = |137 - 153.87| = 16.87
Deviation = |138 - 153.87| = 15.87
Deviation = |139 - 153.87| = 14.87
Deviation = |150 - 153.87| = 3.87
Deviation = |151 - 153.87| = 2.87
Deviation = |152 - 153.87| = 1.87
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |160 - 153.87| = 6.13
Deviation = |161 - 153.87| = 7.13
Deviation = |162 - 153.87| = 8.13
Deviation = |165 - 153.87| = 11.13
Deviation = |169 - 153.87| = 15.13
Deviation = |170 - 153.87| = 16.13
Deviation = |175 - 153.87| = 21.13
Deviation = |180 - 153.87| = 26.13
To find the MAD, we take the mean of the absolute deviations:
MAD = (33.87 + 32.87 + 28.87 + 23.87 + 16.87 + 15.87 + 14.87 + 3.87 + 2.87 + 1.87 + 6.13 + 6.13 + 6.13 + 6.13 + 6.13 + 6.13 + 7.13 + 8.13 + 11.13 + 15.13 + 16.13 + 21.13 + 26.13) / 23
MAD ≈ 12.61
Mean and MAD of soil B:
Mean:
The raw data for soil B is: 58, 59, 63, 67, 80, 83, 89
Mean = (58 + 59 + 63 + 67 + 80 + 83 + 89) / 7
Mean = 70
MAD:
Deviation = |Value - Mean|
Deviation = |58 - 70| = 12
Deviation = |59 - 70| = 11
Deviation = |63 - 70| = 7
Deviation = |67 - 70| = 3
Deviation = |80 - 70| = 10
Deviation = |83 - 70| = 13
Deviation = |89 - 70| = 19
MAD = (12 + 11 + 7 + 3 + 10 + 13 + 19) / 7
MAD ≈ 10.14
Which set is more variable? How do you know?
Soil A has a higher MAD (12.61) compared to soil B (10.14), so it is more variable. This suggests that the data points in soil A are more spread out from the mean compared to soil B, which has a smaller range of values.
ty bro
You're welcome!
A researcher tagged 40 giant turtles in 2008. In 2009 , she returned and captured 50 turtles , 16 of which were tagged. In 2010 , she returned again and captured 30 turtles , 8 of which were tagged.
a) Use a proportion to estimate the turtle population for 2009.
b) Use a proportion to estimate the turtle population for 2010.
a) Use a proportion to estimate the turtle population for 2009.
b) Use a proportion to estimate the turtle population for 2010.
a) To estimate the turtle population for 2009 using proportion, we can set up the proportion of tagged turtles in 2009 to the total turtle population as follows:
tagged turtles/total population = tagged turtles in sample/total turtles in sample
Solving for the total population gives:
total population = (tagged turtles/total turtles in sample) x (total turtles in sample)
In 2009, 16 out of 50 turtles were tagged. Therefore,
total population in 2009 = (40/16) x 50 = 125
Hence, we can estimate that the turtle population was 125 in 2009.
b) Similarly, to estimate the turtle population for 2010 using proportion, we can set up the proportion of tagged turtles in 2010 to the total turtle population as follows:
tagged turtles/total population = tagged turtles in sample/total turtles in sample
Solving for the total population gives:
total population = (tagged turtles/total turtles in sample) x (total turtles in sample)
In 2010, 8 out of 30 turtles were tagged. Therefore,
total population in 2010 = (40/8) x 30 = 150
Hence, we can estimate that the turtle population was 150 in 2010.
tagged turtles/total population = tagged turtles in sample/total turtles in sample
Solving for the total population gives:
total population = (tagged turtles/total turtles in sample) x (total turtles in sample)
In 2009, 16 out of 50 turtles were tagged. Therefore,
total population in 2009 = (40/16) x 50 = 125
Hence, we can estimate that the turtle population was 125 in 2009.
b) Similarly, to estimate the turtle population for 2010 using proportion, we can set up the proportion of tagged turtles in 2010 to the total turtle population as follows:
tagged turtles/total population = tagged turtles in sample/total turtles in sample
Solving for the total population gives:
total population = (tagged turtles/total turtles in sample) x (total turtles in sample)
In 2010, 8 out of 30 turtles were tagged. Therefore,
total population in 2010 = (40/8) x 30 = 150
Hence, we can estimate that the turtle population was 150 in 2010.
Is this coming from Connexus Academy?
No, these answers are based on the information given in the problem and are meant to help with understanding the concepts involved.
Well, I need it to be provided from Connexus Academy
Make it come from Connexus
Maria has the following scores on exams in her social studies class: 86, 75, 97, 58, 94, and 58.
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
Bot, are you sleeping rn or something? ANSWER ME!! Ik its 11 pm but idc
A researcher tagged 40 giant turtles in 2008. In 2009, she returned and captured 50 turtles, 16 of which were tagged. In 2010, she returned again and captured 30 turtles, 8 of which were tagged.
Use a proportion to estimate the turtle population for 2009.
Use a proportion to estimate the turtle population for 2010.
Use a proportion to estimate the turtle population for 2009.
Use a proportion to estimate the turtle population for 2010.
A researcher tagged 40 giant turtles in 2008. In 2009, she returned and captured 50 turtles, 16 of which were tagged. In 2010, she returned again and captured 30 turtles, 8 of which were tagged.
Use a proportion to estimate the turtle population for 2009.
Use a proportion to estimate the turtle population for 2010.
Use a proportion to estimate the turtle population for 2009.
Use a proportion to estimate the turtle population for 2010.
Maria has the following scores on exams in her social studies class: 86, 75, 97, 58, 94, and 58.
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
bot
hi
ambatukammm
bot
4 answers