Multiple choices | Mathematics homework help
1. In a poll, respondents were asked whether they had constantly been in a car clothing. 220 respondents indicated that they had been in a car clothing and 370 respondents said that they had not been in a car clothing. If one of these respondents is fortuityly separated, what is the likelihood of getting someone who has been in a car clothing? Round to the unswerving thousandth.A. 0.384 B. 0.380 C. 0.373D. 0.370
P(accident) = 220 / (220 + 370) = 0.373
2. If you flip a invent three eras, the likely issues are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. What is the likelihood of getting at averageest one crown?
A. 4/9 B. 5/6 C. 7/8 D. 5/8
P(at averageest one crown) = (Number of unions delay one or over H) / (Total enumerate of unions)
P(at averageest one crown) = 7 / 8
3. Joe negotiatet 20 cards from a banner 52-card marshal, and the enumerate of red cards exceeded the enumerate of sombre cards by 8. He reshuffled the cards and negotiatet 30 cards. This era, the enumerate of red cards exceeded the enumerate of sombre cards by 10. Individualize which negotiate is closer to the 50/50 association of red/sombre wait-fored of untarnishedly negotiatet hands from a untarnished marshal and why.
A. The pristine sequence is closer accordingly 1/10 is farther from 1/2 than is 1/8.
B. The sequence closer to the presumptive 50/50 cannot be stable reasonable the enumerate of red and sombre cards for each negotiate is ardent.
C. The assist sequence is closer accordingly 20/30 is closer to 1/2 than is 14/20.
D. The pristine sequence is closer accordingly the estrangement betwixt red and sombre is trivialer than the estrangement in the assist sequence.
1st negotiate: 14 red and 6 sombre P(red) = 14/20 = 0.70
2nd negotiate: 20 red and 10 sombre P(red) = 20/30 = 0.67 < 0.70
4. Suppose you possess an extremely ununtarnished die: The likelihood of a 6 is 3/8, and the likelihood of each other enumerate is 1/8. If you throw the die 32 eras, how numerous twos do you wait-for to see?A. 2 B. 4 C. 3 D. 5
E(2) = np = 32(1/8) = 4
5. A assort consists of 50 women and 82 men. If a tyro is fortuityly separated, what is the likelihood that the tyro is a mother?
A. 32/132 B. 25/66C. 50/132 D. 82/132
P(woman) = 50 / (50 + 82) = 50/132 = 25/66
6. If you flip a invent three eras, the likely issues are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. What is the likelihood that at averageest two crowns supervene orderlyly?
A. 1/8 B. 3/8 C. 5/8 D. 6/8
P(two orderly H) = (Number of unions delay HH) / (Total enumerate of combinations)
P(two orderly H) = 3/8
7. Suppose you buy 1 ticket for $1 out of a lottery of 1000 tickets where the ovation for the one seductive ticket is to be $500. What is your wait-fored treasure?
A. $0.00 B. −$0.40 C. −$1.00 D. −$0.50
E(x) = (1/1000)($500) – 1 = $0.50 - $1 = -$0.50
8. If you flip a invent three eras, the likely issues are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. What is the likelihood of getting at averageest two tails?
A. 1/2 B. 2/3 C. 3/4 D. 4/9
P(at averageest 2 T) = (Number of unions delay 2 or over T) / (Total enumerate
P(at averageest 2 T) = 4/8 = ½
9. A bag contains lewd remnants of which one is red, one is sky sky cerulean-colored-colored, one is bleak, and one is yellow. A remnant is separated at fortuity from the bag and then replaced in the bag. A assist remnant is then separated at fortuity. Execute a schedule of the likely issues (for pattern, RB plays the issue red remnant followed by sky sky sky sky cerulean-colored-colored-colored remnant) and use your schedule to individualize the likelihood that the two remnants separated are the same hue. (Hint: There are 16 likely issues.)A. 1/4 B. ¾ C. 2/16 D. 3/16
Possible issues: RR RB RG RY
BR BB BG BY
GR GB GG GY
YR YB YG YY
P(two of same hue) = (Number of unions delay same hue) / (Total
enumerate of unions)
P(two of same hue) = 4/16 = ¼
10. A committee of three nation is to be formed. The three nation conquer be separated from a schedule of five likely committee limbs. A unadorned fortuity pattern of three nation is captured, delayout regaining, from the clump of five nation. Using the letters A, B, C, D, E to play the five nation, schedule the likely patterns of bulk three and use your schedule to individualize the likelihood that B is interjacent in the pattern. (Hint: There are 10 likely patterns.)A. 0.6 B. 0.4 C. 0.7 D. 0.8
Sample space: ABC ABD ABE ACD ACE
ADE BCD BCE BDE CDE
P(union includes B) = (Number of unions including B) / (Total
enumerate of unions)
P(union includes B) = 6/10 = 0.6
11. In the pristine sequence of rolls of a die, the enumerate of odd enumerates exceeded the enumerate of equal enumerates by 5. In the assist sequence of rolls of the same die, the enumerate of odd enumerates exceeded the enumerate of equal enumerates by 11. Individualize which sequence is closer to the 50/50 association of odd/equal wait-fored of a untarnishedly rolled die.
A. The assist sequence is closer accordingly the estrangement betwixt odd and equal enumerates is senior than the estrangement for the pristine sequence.
B. The pristine sequence is closer accordingly the estrangement betwixt odd and equal enumerates is near than the estrangement for the assist sequence.
C. Since 1/2 > 1/5 > 1/11, the pristine sequence is closer.
D. The sequence closer to the presumptive 50/50 cannot be stable reasonable the whole enumerate of rolls for twain sequence is ardent.
12. Jody checked the region 12 eras on Monday, and the ultimate digit of the region was odd six eras over than it was equal. On Tuesday, she checked it 18 eras and the ultimate digit was odd eight eras over than it was equal. Individualize which sequence is closer to the 50/50 association of odd/equal wait-fored of such a sequence of region checks.
A. The Monday sequence is closer accordingly 1/6 is closer to 1/2 than is 1/8.
B. The Monday sequence is closer accordingly 6/12 is closer to 0.5 than is 8/18.
C. The Tuesday sequence is closer accordingly the 13/18 is closer to 0.5 than is 9/12.
D. The sequence closest to the presumptive 50/50 cannot be stable delayout penetrating the enumerate of odds and equals in each sequence.
Monday: 9 odd, 3 equal P(odd) = 9/12 = 0.75
Tuesday: 13 odd, 5 equal P(odd) = 13/18 ≈ 0.722 < 0.75
13. Suppose you pay $1.00 to roll a untarnished die delay the intelligence that you conquer get tail $3.00 for rolling a 5 or a 2, pin otherwise. What is your wait-fored treasure?
A. $1.00 B. $0.00 C. $3.00 D. −$1.00
E(x) = (1/3)($3) - $1 = $1 - $1 = $0
14. On a multiple excellent trial, each inquiry has 6 likely exculpations. If you execute a fortuity suspect on the pristine inquiry, what is the likelihood that you are reform?
A. 1/5 B. 1/6 C. 1/4 D. 2/5
P(correct) = Enumerate of reform exculpations / Enumerate of likely excellents
P(correct) = 1/6
15. A consider of two types of exterminate killers was produced on two same exterminate frames. One exterminate killer killed 15% over exterminates than the other. This estrangement was telling at the 0.05 roll. What does this average?
A. The proficiency was due to the deed that there were over exterminates in one consider.
B. The likelihood that the estrangement was due to fortuity over is senior than 0.05.
C. The likelihood that one exterminate killer effected rectify by fortuity over is near than 0.05.
D. There is not ample counsel to execute any blank.
16. The classification of B.A. rates acquired by a topical garden is scheduleed under, by elder.
Liberal Arts 1358
What is the likelihood that a fortuityly separated rate is not in Business?
A. 0.7800 B. 0.8200 C. 0.8300 D. 0.9200
P(not in Business) = (Number not in Business) / Whole enumerate of rates
P(not in Business) = (9313 – 1676) / 9313
P(not in Business) = 0.8200
17. A 28-year-old man pays $125 for a one-year career insurance cunning delay coverage of $140,000. If the likelihood that he conquer feed through the year is 0.9994, to the unswerving dollar, what is the man’s wait-fored treasure for the insurance cunning?
A. $139,916 B. −$41 C. $84 D. −$124
E(x) = (1 – 0.9994)($140,000) - $125 = -$41
18. A bag contains 4 red marbles, 3 sky sky sky sky cerulean-colored-colored-colored marbles, and 7 bleak marbles. If a marble is fortuityly separated from the bag, what is the likelihood that it is sky sky cerulean-colored-colored?
A. 2/11 B. 3/11 C. 5/14 D. 3/14
P(blue) = Enumerate of sky sky sky sky cerulean-colored-colored-colored marbles / Whole enumerate of marbles
P(blue) = 3 / (4 + 3 + 7) = 3/14
19. The grounds set plays the allowance rolls of the limbs of a dominion club. Revere the likelihood that a fortuityly separated limb earns at averageest $98,000.
112,000 126,000 90,000 133,000 94,000 112,000 98,000 82,000 147,000 182,000 86,000 105,000
140,000 94,000 126,000 119,000 98,000 154,000 78,000 119,000
A. 0.4 B. 0.6 C. 0.66 D. 0.7
P(stipend ≥ $98,000) = Enumerate of salaries ≥ 98,000 / Whole enumerate of salaries
P(stipend ≥ $98,000) = 14/20 = 0.7
20. A consider of tyros preamble Statistics 101 was produced. Lewd hundred tyros who learned for over than 10 hours averaged a B. Two hundred tyros who learned for near than 10 hours averaged a C. This estrangement was telling at the 0.01 roll. What does this average?
A. The likelihood that the estrangement was due to fortuity over is senior than 0.01.
B. There is near than a 0.01 fortuity that the pristine clump’s removes were rectify by fortuity over.
C. The proficiency was due to the deed that over nation learned.
D. There is not ample counsel to execute any blank
21. Sample bulk = 400, pattern average = 44, pattern banner flexion = 16. What is the extremity of untruth?
A. 1.4 B. 1.6 C. 2.2 D. 2.6
Assuming a 95% reliance roll, and noting that this assort seems to use z = 2 for this reliance roll instead of the over deferential 1.96 treasure.
E = z*s/√n = 2 * 16 / √400 = 1.6
22. Select the best fit frequentedion on the plant diagram under.
D. None of the frequentedions is the frequentedion of best
23. A researcher wishes to revere the average total of capital departed per month on prop by households in a actual neighborhood. She desires a extremity of untruth of $30. Past studies hint that a population banner flexion of $248 is cool. Revere the insufficiency pattern bulk needed to revere the population average delay the methodic exactness.
A. 274 B. 284 C. 264 D. 272
Again, grand a 95% reliance cessation, delay z = 2.00 instead of 1.96:
n = (z * s / E)2 = (2 * 248 / 30)2 = 273.35 à 274
24. Suggest the object of the apposition inchoate the grounds.
The graph demonstrations force of coffee (y) and enumerate of scoops used to execute 10 cups of coffee (x). Identify the reasonable object of the apposition.