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adell [148]
3 years ago
10

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!Savannah recently bought a 65 inch flat screen television. Television dimensions refer to the diagonal size of the screen. Savannah knows that the television is 56.7 inches wide. The height of the television is approximately inches.
Mathematics
2 answers:
aliya0001 [1]3 years ago
6 0

Answer:

Approximately 32 in

Step-by-step explanation:

This is a Pythagorean Theorem problem: we have the hypotenuse (diagonal size) and the leg (width). Now solve for the height.

a^2 + b^2 = c^2

56.7^2 + b^2 = 65^2

b^2 = 1010.11

<u>b = 31.78 in ≈ 32 in</u>

goldfiish [28.3K]3 years ago
4 0

Answer:

. The height of the television is approximately  32 inches or 2 1/3 feet

Step-by-step explanation:

Use Pythagoras' Theorem that says that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.

a²+b² = c²

diagonal =c = 65

56.7 could be a or b so

56.7²+b² = 65²

3124.89+ b² = 4225

b² = 4225 - 3124.89

b² = 1010.11 take a √ to find b

b = 31.78

round to 32 inches

1 foot =12 inches

32/12 = 8/3 = 2 1/3

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(a) The probability that each of them win five games is 0.2461.

(b) The probability that the computer wins seven games is 0.1172.

(c) The probability that the human wins at least 7 games is 0.1711.

Step-by-step explanation:

Let the random variable <em>X</em> = the human wins a chess game.

The probability that the human wins a game is, P (X) = <em>p</em> = 0.50.

The number of games played by the computer and the human is, <em>n </em>= 10.

The random variable X\sim Bin(n = 10,p=0.50)

The probability distribution of the Binomial random variable <em>X</em> is:

P (X=x)={n\choose x}p^{x}(1-p)^{n-x}={10\choose x}(0.50)^{x}(1-0.50)^{10-x}

(a)

If both the computer and the human wins five games each then the probability of the human winning 5 games is:

P (X=5)={10\choose 5}(0.50)^{5}(1-0.50)^{10-5}\\=252\times 0.03125\times0.03125\\=0.246094\\\approx 0.2461

Thus, the probability that each of them win five games is 0.2461.

(b)

If the computer wins 7 games then the number of games won by the human is, 10 - 7 = 3.

P (Computer winning 7 games) = P (Human winning 3 games)

The probability that the human wins 3 games is:

P (X=3)={10\choose 3}(0.50)^{3}(1-0.50)^{10-3}\\=120\times 0.125\times0.0078125\\=0.1171875\\\approx 0.1172

Thus, the probability that the computer wins seven games is 0.1172.

(c)

Compute the probability that the human wins at least 7 games as follows:

P(X\geq 7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)\\={10\choose 7}(0.50)^{7}(1-0.50)^{10-7}+{10\choose 8}(0.50)^{8}(1-0.50)^{10-8}\\+{10\choose 9}(0.50)^{9}(1-0.50)^{10-9}+{10\choose 10}(0.50)^{10}(1-0.50)^{10-10}\\=0.1172+0.044+0.009+0.0009\\=0.1711

Thus, the probability that the human wins at least 7 games is 0.1711.

5 0
3 years ago
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