Hey guys , Can someone help with the following problems: 1)The residents of town x participated in a survey to determine no.of. hrs per week each resident spent watchin tv. The distribution of results of survey had a mean of 21 hrs and a std deviation of 6 hrs. the no. of hrs that Pat a resident of town x watch tv last week was between 1 & 2 std deviation bvelow mean. Which of the foll could be a no. of hrs that Pat saw TV last week. 30 20 18 12 6 2) In the xy plane, line k has positive slope and x intercept is 4. If the area of the triangle formed by line k and the two axes is 12, what is the y intercept of line k? 1)-6 2)-4 3)-3 4)3 5)6 3) Which of the following is equal to (2^12 2^6) / (2^6 2^3)? A. 2^6 + 2^3 B. 2^6 2^3 C. 2^9 D. 2^3 E. 2
Also, need to know if there is a specific way to solve the problems on graphs i.e findind the x or y intercept of a line equation.
Thanx!
hi, Q1) 1 standard deviation below mean is 21 -6 = 15. 2 standard deviations below mean is 21 -2*6 = 9. So Pat's hours must be between 9 and 15. Only 12 falls in this range. Hence answer is D.
Q2)Slope is positive.x-intercept is +4. Hence the triangle formed by line k and the axes must fall in the 4th quadrant. Since x-intercept is 4, (4,0) falls on line k.Area is 12. If y is the y-intercept, (4*y)/2 = 12. So y = 6. Since triangle is in the 4th quadrant, y-intercept must be negative. Hence -6 is the answer.A.
Q3) (2^12 - 2^6) / (2^6 - 2^3) can be simplified to {2^6 ( 2^6 - 1)} / {2^3 ( 2^3 -1)} or 2^3 * 9 = 72. Option A = 72. So A is correct.
Hey guys , Can someone help with the following problems: 1)The residents of town x participated in a survey to determine no.of. hrs per week each resident spent watchin tv. The distribution of results of survey had a mean of 21 hrs and a std deviation of 6 hrs. the no. of hrs that Pat a resident of town x watch tv last week was between 1 & 2 std deviation bvelow mean. Which of the foll could be a no. of hrs that Pat saw TV last week. 30 20 18 12 6 2)In the xy plane, line k has positive slope and x intercept is 4. If the area of the triangle formed by line k and the two axes is 12, what is the y intercept of line k? 1)-6 2)-4 3)-3 4)3 5)6 3) Which of the following is equal to (2^12 2^6) / (2^6 2^3)? A. 2^6 + 2^3 B. 2^6 2^3 C. 2^9 D. 2^3 E. 2 Also, need to know if there is a specific way to solve the problems on graphs i.e findind the x or y intercept of a line equation. Thanx!
1) Ans : 12 between 1 and 2 st deviations below mean implies range is between 9 and 15..consider the ans options..only 12 is possible...
2) Ans: -6 positive slope implies line makes an acute inclination wrt to postive X - Axis... and X intercept is 4 implies line cuts X axis at 4 or passes thru pt (4,0)... now, +ve X intercept and +ve slope implies y intercept is -ve ... And area is 12.. Hence Y intercept is -6 ( 1/2*6*4) = 12
Also....... need to know if there is a specific way to solve the problems on graphs i.e findind the x or y intercept of a line equation.
You know the equation of a line: y=mx+c. To get y-intercept, substitute x = 0. You get c. To get x-intercept, substitute y = 0. You get -c/m.
But you need to see what data is given in the problem.This is a very basic way of finding the intercepts.For the problem you posted, I already told u how to find the y-intercept.
A committee of three people is to be chosen from four married couples. What is the number of different committees can be chosen if two ppl who are married to each other cannot serve on the committee?
1) [LEFT]A string of 10 lightbulbs is wired in such a way that if any individual lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during time period T id 0.06, what is the probability that the string of lightbulbs will fail during time period T? A.0.06 B.(0.06)^10 C.1-(0.06)^10 D.(0.94)^10 E.1-(0.94)^10
2)[/LEFT] [LEFT]In the xy-plane, if line k passes through the points (3.-4) and (a,b), where b=4a-16 and a􀂘3, what is the slope of k? A. 4 B. 1/2 C. 0 D. 2 E. 4
3) Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?[/LEFT] [LEFT]A. 24 B. 36 C. 72 D. 144 E. 216[/LEFT]
A committee of three people is to be chosen from four married couples. What is the number of different committees can be chosen if two ppl who are married to each other cannot serve on the committee?
a)16 b)24 c)26 d)30 e)32
is d ans 32 ??
first person can be chosen in 8 ways.. 2nd person can be chosen in 6 ways.. 3rd person can be chosen in 4 ways..
total ways = 8*6*4 order is not imp, we just have to select.. total ways of forming committee is 8*6*4/3! = 32 ways..
3) Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two? A. 24 B. 36 C. 72 D. 144 E. 216
Consider one case:
11_ _ could be anything from 2 to 9.
Hence, 8 such cases. Similarly, 8 each for 1_1 and _11.
Total for 1 = 24 There are 9 such numbers. Hence, 9*24 = 216.
1) [LEFT]A string of 10 lightbulbs is wired in such a way that if any individual lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during time period T id 0.06, what is the probability that the string of lightbulbs will fail during time period T? A.0.06 B.(0.06)^10 C.1-(0.06)^10 D.(0.94)^10 E.1-(0.94)^10[/LEFT]
[LEFT]2) In the xy-plane, if line k passes through the points (3.-4) and (a,b), where b=4a-16 and a􀂘3, what is the slope of k? A. 4 B. 1/2 C. 0 D. 2 E. 4[/LEFT]
[LEFT]3) Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two? A. 24 B. 36 C. 72 D. 144 E. 216[/LEFT]
1) p(string failing) = 1- p(string not failing )
p(string not failing) = p ( none of the bulbs fail
Hence, p( string failing) = 1 - 0.94^10...ANS E
2) question not clear, what is a􀂘3 3) 1st digit can be chosen in 9 ways.. 2nd digit can be chosen in 1 way 3rd digit can be chosen in 8 ways...
total ways = 9*1*8= 72 ways..and 3 chosen digits can be arranged among themselves such that 2 digits are same and 3rd is different in 3!/2 = 3 ways..
A committee of three people is to be chosen from four married couples. What is the number of different committees can be chosen if two ppl who are married to each other cannot serve on the committee?
a)16 b)24 c)26 d)30 e)32
hi Bhavin,
yes 32 is the answer.I was so frustrated that i did not get the answer from
yes 32 is the answer.I was so frustrated that i did not get the answer from
8 X 6 X 4.
what is the formula for 8 X 6 X 4/3! ?
Thanks, raj
ok raj...let me explain..there are 4 couples...let them be A1, A2, B1, B2, C1, C2, D1, D2
we have to choose 3 ppl, such that none of them are couples..
1st person can be any 1 of the 8 ppl...hence 8 possibilities..let us say A1 is chosen.. now 7 ppl are left, we can choose any 1 of them except the spouse (A2)
Hence, 2nd person can be any 1 of the remaining 6 ppl...hence 6 possibilities for 2nd spot..let us say B1 is chosen.. now 6 ppl are left, but we cannot choose spouses of previous 2 selected candidates (A2 & B2 cannot be selected )
Hence, 3rd person can be any 1 of the remaining 4 ppl...hence 4 possibilities for 3rd slot..let us say C1 is chose..
Hence, we have 8*6*4 total possibilities...but we do not have to arrange them fr a photograph or smtng...hence order is not imp...hence u divide by the no of ways they can be mutually arranged..ie 3! ways..
i.e A1, B1, C1 OR B1, A1,C1 and so on is not imp..
A committee of three people is to be chosen from four married couples. What is the number of different committees can be chosen if two ppl who are married to each other cannot serve on the committee?
a)16 b)24 c)26 d)30 e)32
One more approach from my side:
Well from each couple we can pick only one: so lets pick 3 couples first, in 4C3 i.e. 4 ways. Now from each of these couple we can choose 1 in 2 ways. So the answer is 4C3*2^3 = 32.
p(string not failing) = p ( none of the bulbs fail
Hence, p( string failing) = 1 - 0.94^10...ANS E
2) question not clear, what is a􀂘3
3) 1st digit can be chosen in 9 ways.. 2nd digit can be chosen in 1 way 3rd digit can be chosen in 8 ways... total ways = 9*1*8= 72 ways..and 3 chosen digits can be arranged among themselves such that 2 digits are same and 3rd is different in 3!/2 = 3 ways.. total poss 3 digit nos = 72*3=216 ways.. hope this solves your doubt..
Bhavin,
In the first question, isnt the prob of failure of each light bulb =.06 ? And hence, for the string of 10bulbs, Prob(failure of string)=.06^10 .
In the third question, I have used another approach.
Total possible 3 digit numbers without the digit zero = 9*9*9 = 729 The no of 3 digit nos with different digits = 9*8*7 = 504 We have 9 numbers with all the 3 digits same.
Hence the no. of 3 digit numbers having 2 digits same and the third digit different = 729-(504+9) = 216.
In the first question, isnt the prob of failure of each light bulb =.06 ? And hence, for the string of 10bulbs, Prob(failure of string)=.06^10 .
Hi kingcat,
statement in bold is not correct.. problem states that string fails if any of the bulb fails..
0.06^10 is when all bulbs fail...and that is not the criteria for string failing.. you have mulitple poss as to how a string can fail.. say, bulb 1 fails and other 9 bulbs dont fail
OR
bulb 2 fails and other 9 bulbs dont fail..
and many more..this is tedious...so u consider the counterpart.. find the chances of none of the bulbs failing, hence the string would not fail and subtract it from 1..
p( each bulb fails ) =0.06 Hence p(bulb does not fail= 1-0.06=0.94 Hence p(none of the bulbs fail) = 0.94^10 Hence p(string does not fail) = 0.94^10 Hence p(string fails) = 1-p(string does not fail) 1-0.94^10
Hi Bhavin , Two more problems: 1)Sum of 50 positive integers is 2550. What is he sum of even integers from 102 to 200 inclusive? 5100 7550 10100 15500 20100
2) Amys grade was the 90th percentile of he 80 grades for her class. Of the 100 grades from another class, 19 were higher than amys and the rest were lower. If no other grade was same as amys grade then amys grade was what percntile of the grades of he two classes combined. 72nd 80th 81st 85th 92nd.
1)Sum of 50 positive integers is 2550. What is he sum of even integers from 102 to 200 inclusive? 5100 7550 10100 15500 20100
Sum of 50 positive integers is 2550.: This info is incomplete. It should be sum of first 50 even integers is 2550 Then the question can be solved very quickly without any deep calculations as: sum (102,104,106..200) = sum ( 2,4,6..100) + sum (100 + 100 +100 .. + 100 (50 terms)) = 2550 (given) + 100*50
Because the first information was incomplete, and I didn't care to verify it, I got 10100 as answer.:oops: