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saveliy_v [14]
3 years ago
8

There are 10 cards. Each card has one number between 1 and 10, so that every number from 1 to 10 appears once.

Mathematics
1 answer:
pychu [463]3 years ago
6 0

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:


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Which statement must be true to be able to use the AAS Congruence Theorem to prove triangle LMN is congruent to triangle PON?
aev [14]

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a

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2 years ago
ACCTUALLY HELP ME??!!!!!!! NEEDED FOR TOMORROW AND 50 POINTS! will award brainliest for first answer!!
drek231 [11]

Answer:

a:c = 35:24

a:c = 20:27

a:c = 35:22

a:c = 28:27

Step-by-step explanation:

a:b = 7:3

Using cross products

3a = 7b

Divide by 7

3a/7 = b

Now we want

8b = 5c

Substitute in 3a/7 for b

8 (3a/7) = 5c

24/7a = 5c

Multiply by 7

24/7a *7 = 5*7c

24a = 35c

Divide by c

24 a/c = 35

Divide by 24

a/c = 35/24

a:c = 35:24

a:b = 4:9

Using cross products

9a = 4b

Divide by 4

9a/4 = b

Now we want

3b = 5c

Substitute in 9a/4 for b

3 (9a/4) = 5c

27/4a = 5c

Multiply by 4

27/4a *4 = 5*4c

27a = 20c

Divide by c

27 a/c = 20

Divide by 27

a/c = 20/27

a:c = 20:27

b:c = 5:11

Using cross products

11b = 5c

Divide by 11

b = 5c/11

Now we want

2a = 7b

Substitute in 5c/11 for b

2a = 7(5c/11)

2a = 35c/11

Multiply by 11

2a*11 = 35c

22a = 35c

Divide by c

22 a/c = 35

Divide by 22

a/c = 35/22

a:c = 35:22

b:c = 14:3

Using cross products

3b = 14c

Divide by 3

b = 14c/3

Now we want

9a = 2b

Substitute in 14c/3 for b

9a = 2(14c/3)

9a = 28c/3

Multiply by 3

9a*3 = 28c

27a = 28c

Divide by c

27 a/c = 28

Divide by 27

a/c = 28/27

a:c = 28:27

7 0
3 years ago
Find an explicit solution of the given initial-value problem. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0
MissTica

Answer:

a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x*(1 + 4y²) dx = 0

(1 + x⁴) dy  = - x*(1 + 4y²) dx

[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx

∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx

now to solve each integral

I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁

I₂=  ∫[-x/(1 + x⁴)] dx

for u= x² → du=x*dx

I₂=  ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ =  - tan⁻¹ (x²) +C₂

then

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C

for y(x=1) = 0

1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C

since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for  π*N , we will choose for simplicity N=0 . hen an explicit solution would be

1/2 * 0 = - π/4 + C

C= π/4

therefore

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

5 0
3 years ago
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