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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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eimsori [14]

Answer:

option c

x + 4 = 64

and

option e

x = 60  

Step-by-step explanation:

Given in the question an expression

3∛x+4 = 12

∛x + 4 = 12/3

∛x + 4 = 4

Take cube on both sides of the equation

∛(x + 4)³ = 4³

x + 4 = 64

x = 64 - 4

x = 60  

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Step-by-step explanation:

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3 years ago
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Answer:

a) [-0.134,0.034]

b) We are uncertain

c) It will change significantly

Step-by-step explanation:

a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.

Since we assume that the variances are equal, we use the pooled variance given as

s_p^2 = \frac{ (n_1 -1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2},

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The confidence interval is

(\mu_1 - \mu_2) \pm t_{n_1+n_2-2,\alpha/2} \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} = (-0.05) \pm t_{68,0.025} \sqrt{\frac{0.03}{40} + \frac{0.03}{30}}

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4 years ago
How many numbers are equal to the sum of two odd, one digit numbers
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Here is a list of the odd number paired

1+3, 1+5, !+7, and !+9  (there are 4 unique sums - 4, 6,8 and 10)
3+5, 3+7, 3+9  (notice I did not pair 3 with 1 and the the only new sum is 12)
5+7, 5+9  (the only new sum is 14)
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The sums (no repeats) are 4,6,8,10,12,14 and 16 for a total of seven numbers.
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actually there are two cases, don't have intersection and have. if have intersection, then they intersect at line y = x or point (a, a) by definition of inverse function.

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