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Mathematics 12 Online
OpenStudy (anonymous):

Weights of male mountain lions follow the normal distribution with a median of 150 lb and an interquartile range of 8.2 lb.

OpenStudy (anonymous):

what's the question, could you be more specific?

OpenStudy (anonymous):

Find the 75th percentile of the weights

OpenStudy (anonymous):

a.154.11 b.160

OpenStudy (anonymous):

0.674*6.12 + 150 = 154.11 1.645*6.12 + 150 = 160

OpenStudy (anonymous):

Undergraduates in a large university class were asked to record the time they spent on the most recent homework. In a histogram of the recorded times, the height of the bar over the interval “1 hour to 3 hours” is 4% per hour. The height of the bar over the interval “3 hours to 6 hours” is 2% per hour. If the histogram is redrawn with just a single bar over the interval “1 hour to 6 hours”, what should the height of the bar be?

OpenStudy (anonymous):

@ankitojha, please can u explain to me how you got this correct answer?

OpenStudy (anonymous):

0.674*6.12 + 150 = 154.11 1.645*6.12 + 150 = 160

OpenStudy (anonymous):

@ankitojha, @dinakar, i mean how did u convert the iqr 8.2 to 6.12

OpenStudy (anonymous):

dnt knw exactly, ill let u knw

OpenStudy (anonymous):

Undergraduates in a large university class were asked to record the time they spent on the most recent homework. In a histogram of the recorded times, the height of the bar over the interval “1 hour to 3 hours” is 4% per hour. The height of the bar over the interval “3 hours to 6 hours” is 2% per hour. If the histogram is redrawn with just a single bar over the interval “1 hour to 6 hours”, what should the height of the bar be?

OpenStudy (anonymous):

Q3-Q1= SD(z3-z1) 8.2 = SD(0.67-(-0.67)) SD=8.2/1.34 SD=6.12 0.674*6.12 + 150 =154.11 for 75th percentile 1.645*6.12 + 150 = 160 for 95th percentile

OpenStudy (anonymous):

Undergraduates in a large university class were asked to record the time they spent on the most recent homework. In a histogram of the recorded times, the height of the bar over the interval “1 hour to 3 hours” is 4% per hour. The height of the bar over the interval “3 hours to 6 hours” is 2% per hour. If the histogram is redrawn with just a single bar over the interval “1 hour to 6 hours”, what should the height of the bar be?

OpenStudy (anonymous):

http://en.wikipedia.org/wiki/Interquartile_range

OpenStudy (anonymous):

@ankitojha, you ROCK!

OpenStudy (anonymous):

can u reply for my new question

OpenStudy (anonymous):

The scatter diagram of the lengths and widths of the petals of a type of flower is football shaped; the correlation is 0.8. The average petal length is 3.1 cm and the SD is 0.25 cm. The average petal width is 1.8 cm and the SD is 0.2 cm.

OpenStudy (anonymous):

One of the petals is 0.2 cm wider than another. Regression says that the wider petal is estimated to be ____________ cm longer than the other. unanswered

OpenStudy (anonymous):

The scatter diagram of midterm scores and final scores in a large class is football shaped. For about 80% of the students, the regression estimate of final score based on midterm score is correct to within 15 points. For about 95% of the students, the regression estimate of final score based on midterm score is correct to within ___________ points.

OpenStudy (anonymous):

can any one pls ans the above questions

OpenStudy (anonymous):

6B : 1.5-1.8/0.2=-1.2 -1.2*0.25+3.1 =2.8 Ans :2.8

OpenStudy (anonymous):

Find the regression estimate of the length of a petal that is 1.5 cm wide. 1.5-1.8/0.2=-1.2 -1.2*0.25+3.1 =2.8 Ans :2.8

OpenStudy (anonymous):

have u done with other questions ,if so can u post the answers

OpenStudy (anonymous):

can u please post the answers for other above questions

OpenStudy (anonymous):

The scatter diagram of midterm scores and final scores in a large class is football shaped. For about 80% of the students, the regression estimate of final score based on midterm score is correct to within 15 points. For about 95% of the students, the regression estimate of final score based on midterm score is correct to within ___________ points.

OpenStudy (anonymous):

can any one answer this above question

OpenStudy (anonymous):

24.7 find the z score value of the 95% on the applet and multiply it by the 15th point of the 80%

OpenStudy (anonymous):

Undergraduates in a large university class were asked to record the time they spent on the most recent homework. In a histogram of the recorded times, the height of the bar over the interval “1 hour to 3 hours” is 4% per hour. The height of the bar over the interval “3 hours to 6 hours” is 2% per hour. If the histogram is redrawn with just a single bar over the interval “1 hour to 6 hours”, what should the height of the bar be?

OpenStudy (anonymous):

can any one answer the above question

OpenStudy (anonymous):

hint: 3-1 = 2*4% =8% 6-3 = 3*2% = 6% add the two resulting %s and divide by 5 which is (6-1) whatever u get is the answer, let me know if it helped

OpenStudy (anonymous):

6D : -0.125*0.25 + 3.1 6E : -0.125*0.15 + 3.1

OpenStudy (anonymous):

can u post the exact answer

OpenStudy (anonymous):

(8%+6%)/5 = 2.8

OpenStudy (anonymous):

ya fne iam working on that .i wil let u know

OpenStudy (anonymous):

Of the petals that are 1.5 cm wide, about what percent are less than 2.75 cm long? [Do not enter the percent sign.]

OpenStudy (anonymous):

can any one answer the above question

OpenStudy (anonymous):

The scatter diagram of the lengths and widths of the petals of a type of flower is football shaped; the correlation is 0.8. The average petal length is 3.1 cm and the SD is 0.25 cm. The average petal width is 1.8 cm and the SD is 0.2 cm. 6c: Of the petals that are 1.5 cm wide, about what percent are less than 2.75 cm long?

OpenStudy (anonymous):

The scatter diagram of the lengths and widths of the petals of a type of flower is football shaped; the correlation is 0.8. The average petal length is 3.1 cm and the SD is 0.25 cm. The average petal width is 1.8 cm and the SD is 0.2 cm. 6c: Of the petals that are 1.5 cm wide, about what percent are less than 2.75 cm long?

OpenStudy (anonymous):

6c: Of the petals that are 1.5 cm wide, about what percent are less than 2.75 cm long?

OpenStudy (anonymous):

The scatter diagram of the lengths and widths of the petals of a type of flower is football shaped; the correlation is 0.8. The average petal length is 3.1 cm and the SD is 0.25 cm. The average petal width is 1.8 cm and the SD is 0.2 cm. 6c: Of the petals that are 1.5 cm wide, about what percent are less than 2.75 cm long?

OpenStudy (anonymous):

The scatter diagram of the lengths and widths of the petals of a type of flower is football shaped; the correlation is 0.8. The average petal length is 3.1 cm and the SD is 0.25 cm. The average petal width is 1.8 cm and the SD is 0.2 cm. 6c: Of the petals that are 1.5 cm wide, about what percent are less than 2.75 cm long?

OpenStudy (anonymous):

37

OpenStudy (anonymous):

6c ans=37

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