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trasher [3.6K]
3 years ago
15

A True/False quiz has three questions. When guessing, the probability of getting a question correct is the same as the probabili

ty of getting a question wrong. What is the probability that a student that guesses gets at least 2 questions correct? (Give your answer to 2 decimal places)
Mathematics
1 answer:
rewona [7]3 years ago
6 0

Answer:

iycdjrasgmbjbljvkg

Step-by-step explanation:

xgcuchjlhgjfa1347722j

You might be interested in
There are 10 cards. Each card has one number between 1 and 10, so that every number from 1 to 10 appears once.
pychu [463]

Answer:

When a card is chosen at random with replacement five times, X is the number of times a prime number is chosen.          Here the card is chosen with replacement.  This implies probability for choosing a prime number remains the same as the previously drawn card is replaced.

The sample space= {1,2,3,4,5,6,7,8,9,10}

Prime numbers = {2,3,5,7}

Prob for drawing prime number = 4/10 = 0.4

is the same when replacement is done.

Also there are two outcomes either prime or non prime.  Hence in this case, X the no of times a prime number is chosen, is binomial with p =0.4 and q = 0.6 and n=5


When a card is chosen at random without replacement three times, X is the number of times an even number is chosen.

Prob for an even number = 0.5

But after one card drawn say odd number next card has prob for even number as 5/9 hence each draw is not independent of the other.  Hence not binomial.

When a card is chosen at random with replacement six times, X is the number of times a 3 is chosen.

Here since every time replacement is done, probability of drawing a 3 remains constant = 1/10 = 0.3

i.e. each draw is independent of the other and there are only two outcomes , 3 or non 3. Hence here X is binomial.

When a card is chosen at random with replacement multiple times, X is the number of times a card is chosen until a 5 is chosen

Here X is the number of times a card is chosen with replacement till 5 is chosen.  This is not binomial.  Here probability for drawing nth time correct 5 is  P(non 5 in the first n-1 draws)*P(5 in nth draw) = 0.1^(n-1) (0.9)

Because nCr is not appearing i.e. 5 cannot appear in any order but only in the last draw, this is not binomial.

Step-by-step explanation:


6 0
3 years ago
Factor the GCF: −12x4y − 9x3y2 + 3x2y3
monitta
Factor the coefficients:
-12=(-1)(3)(2^2)
-9=(-1)(3^2)
3=3

The greatest common factor (GCF) is 3

Next we find the GCF for the variable x.
x^4
x^3
x^2
The GCF is x^2.

Next GCF for variable y.
y
y^2
y^3
the GCF is y

Therefore the GCF is 3x^2y
To factor this out, we need to divide each term by the GCF,
(3x^2y)(−12x4y/(3x^2y) − 9x3y2/(3x^2y) + 3x2y3/(3x^2y) )
=(3x^2y)(-4x^2-3xy+y^2)     
if we wish, we can factor further:
(3x^2y)(y-4x)(x+y)
3 0
3 years ago
HURRY !! which comparison of the two equations is accurate? (picture)
VladimirAG [237]

Answer:

its c .Both equations have the same potential solutions, but equation A might have extraneous solutions.

Step-by-step explanation:

just took the test

7 0
3 years ago
Read 2 more answers
Suppose James randomly draws a card from a standard deck of 52 cards. He then places it back into the deck and draws a second ca
qwelly [4]

Answer:

1.92%

Step-by-step explanation:

The probability for first case, picking a queen out of deck, will be:

\frac{4}{52}

as there will be 4 queens in a deck, one of each suit.

For the second pick, the probability of picking a diamond card, will be:

\frac{13}{52}

here the total will remain 52 as he has replaced the first card and not kept it aside and there will be 13 cards in diamond suit (including the three face cards).

Thus the net probability for both cases will be:

P = \frac{4}{52}  * \frac{13}{52}\\ P = \frac{1}{52}\\ P = 0.01923\\P = 1.923\%\\\\P = 1.92\%

Thus total probability for the combined two cases will be 1.92%

7 0
2 years ago
HELP ASAP!! SOMEONE DO NUMBER 7 AND 8. ILL GIVE BRAINLIEST
bija089 [108]

Answer:

3.6 3/6 3.6 percent 3:6

Step-by-step explanation:

10:9

15:14

20:19

4 0
2 years ago
Read 2 more answers
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