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lord [1]
4 years ago
11

Which number is the best approximate of square root of 500 to the nearest tenth

Mathematics
1 answer:
Vika [28.1K]4 years ago
6 0
The most approximate answer would be 22.36 but rounding you get 22.4 so 22.4
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Which is greater 2/5 or 2/6
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An air force intercept squadron consists of 16 planes that should always be ready for immediate launch. However, a plane's engin
prisoha [69]

Answer:

(a) The mean, variance and standard deviation of planes that would be successfully launched are 12, 3 and 1.732 respectively.

(b) The probability that exactly 12 planes scramble successfully is 0.225.

(c) The probability that at least 14 planes scramble successfully is 0.1971.

Step-by-step explanation:

Let <em>X</em> = number of planes whose engines will start at the first attempt.

The probability of <em>X</em> is:

P (A plane starting immediately) = 1 - P (A plane not starting at the 1st attempt)

                                                     = 1 - 0.25

                                                  <em>p </em>= 0.75

The number of planes at the air force intercept squadron is, <em>n</em> = 16.

The random variable <em>X</em> follows a Binomial distribution, i.e.X\sim Bin(16,0.75)

(a)

The expected value of a Binomial distribution is:

E(X)=np

Compute the expected number of planes that would be successfully launched as follows:

E(X)=np=16\times0.75=12

The variance of a Binomial distribution is:

V(X)=np(1-p)

Compute the variance of planes that would be successfully launched as follows:

V(X)=np(1-p)=16\times0.75\times (1-0.75)=3

Compute the standard deviation of planes that would be successfully launched as follows:

SD(X)=\sqrt{V(X)}=\sqrt{3}=1.732

Thus, the mean, variance and standard deviation of planes that would be successfully launched are 12, 3 and 1.732 respectively.

(b)

The probability function of a Binomial distribution is:

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

Compute the probability that exactly 12 planes scramble successfully as follows:

P(X=12)={16\choose 12}(0.75)^{12}(1-0.75)^{16-12}=1820\times0.0317\times0.0039=0.225

Thus, the probability that exactly 12 planes scramble successfully is 0.225.

(c)

Compute the probability that at least 14 planes scramble successfully as follows:

P (X ≥ 14) = P (X = 14) + P (X = 15) + P (X = 16)

                ={16\choose 14}(0.75)^{14}(1-0.75)^{16-14}+{16\choose 15}(0.75)^{15}(1-0.75)^{16-15}\\+{16\choose 16}(0.75)^{16}(1-0.75)^{16-16}\\=0.1336+0.0535+0.01\\=0.1971

Thus, the probability that at least 14 planes scramble successfully is 0.1971.

3 0
3 years ago
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