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antiseptic1488 [7]
3 years ago
6

Find the Length,width and Height. Find the surface area as well

Mathematics
1 answer:
quester [9]3 years ago
6 0

Answer:

L 12 yd

W 7 yd

and H 10 yd

surface area is 548 sq. yd

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How to solve sets in mathematics ​
FinnZ [79.3K]

Answer:

Sets, in mathematics, are an organized collection of objects and can be represented in set-builder form or roster form. Usually, sets are represented in curly braces {},

For example, A = {1,2,3,4} is a set.

5 0
4 years ago
A total of 2n cards, of which 2 are aces, are to be randomly divided among two players, with each player receiving n cards. Each
klasskru [66]

Answer:

P(X_s^c|X_F) =0.2

P(X_s^c|X_F) =0.31

P(X_s^c|X_F) =0.331

Step-by-step explanation:

From the given information:

Let represent X_F as the first player getting an ace

Let X_S to be the second player getting an ace and

\sim X_S as the second player not getting an ace.

So;

The probabiility of the second player not getting an ace  and the first player getting an ace can be computed as;

P(\sim X_S| X_F) = 1 - P(X_S|X_F)

P(X_S|X_F) = \dfrac{P(X_SX_F)}{P(X_F)}

Let's determine the probability of getting an ace in the first player

i.e

P(X_F) = 1 - P(X_F^c)

= 1 -\dfrac{(^{2n-2}_n)}{(^{2n}_n)}}

= 1 - \dfrac{n-1}{2(2n-1)}

=  \dfrac{3n-1}{4n-2} --- (1)

To determine the probability of the second player getting an ace and the first player getting an ace.

P(X_sX_F) = \text{ (distribute aces to both ) and (select the left over n-1 cards from 2n-2 cards}P(X_sX_F) = \dfrac{2(^{2n-2}C_{n-1})}{^{2n}C_n}

P(X_sX_F) = \dfrac{n}{2n -1}---(2)

P(X_s|X_F) = \dfrac{2}{1}

P(X_s|X_F) = \dfrac{2n}{3n -1}

Thus, the conditional probability that the second player has no aces, provided that the first player declares affirmative is:

P(X_s^c|X_F) = 1- \dfrac{2n}{3n -1}

P(X_s^c|X_F) = \dfrac{n-1}{3n -1}

Therefore;

for n= 2

P(X_s^c|X_F) = \dfrac{2-1}{3(2) -1}

P(X_s^c|X_F) = \dfrac{1}{6 -1}

P(X_s^c|X_F) = \dfrac{1}{5}

P(X_s^c|X_F) =0.2

for n= 10

P(X_s^c|X_F) = \dfrac{10-1}{3(10) -1}

P(X_s^c|X_F) = \dfrac{9}{30 -1}

P(X_s^c|X_F) = \dfrac{9}{29}

P(X_s^c|X_F) =0.31

for n = 100

P(X_s^c|X_F) = \dfrac{100-1}{3(100) -1}

P(X_s^c|X_F) = \dfrac{99}{300 -1}

P(X_s^c|X_F) = \dfrac{99}{299}

P(X_s^c|X_F) =0.331

8 0
3 years ago
HELP!!!! Please it’s 6th grade!!
Mandarinka [93]
It’s the second one with {14, -7, 0, 243, -3}
6 0
4 years ago
Suppose three marksmen shoot at a target. The ith marksman fires ni times, hitting the target each time with probability Pi, ind
astraxan [27]

Answer:

Distribution is NOT binomial.

Step-by-step explanation:

In order to be a binomial distribution, the probability of success for each individual trial must be the same. Since each marksman hits the target with probability Pi, the probability of success (hitting the target) is not necessarily equal for all trials. Therefore, the distribution is not binomial.

In this case, the distribution would only be binomial if Pi was the same for every "ith" marksman.

6 0
3 years ago
POSSIBLE POINTS, 2
Lubov Fominskaja [6]
Jerry purchased 5 corn dogs.
8 0
3 years ago
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