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Anni [7]
2 years ago
11

Assume that an opinion poll conducted in a 1998 congressional race found that on election eve, 54% of the voters supported Congr

essman Stevens and 44% supported challenger Jones. Also assume that the poll had a +/- 3% margin of error. What would the pollster be able to safely predict?
Mathematics
1 answer:
vitfil [10]2 years ago
3 0

Answer:

Congressman Stevens will win the race

Step-by-step explanation:

Considering the margin of error, the possible outcomes for each candidate would be:

Congressman Stevens: from (54 - 3)% to (54+3)%

Challenger Jones: from (44 - 3)% to (44+3)%

Congressman Stevens: from 51% to 57%

Challenger Jones: from 41% to 47%

Therefore, even considering the margin of error, the pollster would be able to safely predict that Congressman Stevens will win the race.

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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
3 years ago
There are 45 new houses being built in a neighborhood. Last month 1/3 of them were sold. This month 2/5 of the remaining houses
polet [3.4K]

Answer:

total=45

1/3+2/5=11/15

remaining fraction of houses=1-11/15

=4/15

reamaimg houses=4/15×45=12houses

7 0
3 years ago
Fraction 4 over 5 n = Fraction 2 over 3. n = ___?
Taya2010 [7]

Answer:

Step-by-step explanation:

(4/5)n = (2/3)

move 4/5 to the right

n = (2/3)/(4/5)

n = 5/6

5 0
3 years ago
Need helpp fastt plss will give brainliest to who is correct plss helpp
alisha [4.7K]
I think I might have my math wrong here but I think 5
8 0
2 years ago
How do I evaluate using the given values <br><br> P+r-2; use p=3 and r=2
natali 33 [55]

Answer:

3

Step-by-step explanation:

P+r-2

=3+2-2

=5-2

=3 is the answer.........................

4 0
3 years ago
Read 2 more answers
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