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3 years ago

0.5(2q+1)=0.75q thanks1

Mathematics

1 answer:

irinina [24]3 years ago

4 0

Answer:

q = -2

Step-by-step explanation:

0.5 x 2q = 1q

0.5 x 1 = 0.5

So you have 1q + 0.5 = 0.75q

Then you subtract, 0.75q - 1q which equals -0.25q = 0.5

You divide -0.25q and get q and 0.5 divided by -0.25 is -2

So, you get q = -2

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Answer:

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(d) $\mu_{X}=6$

(e) $\sigma_{X}=2.45$

Step-by-step explanation:

Let X = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, λ = 6.

The random variable X follows a Poisson distribution with parameter λ = 6.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...$

(a)

Compute the value of P (X = 7) as follows:

$P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388$

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

$=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380$

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

$=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443$

Thus, the value of P (2 < X < 7) is 0.5433.

(d)

The mean of the Poisson distribution is:

$\mu_{X}=\lambda=6$

(e)

The standard deviation of the Poisson distribution is:

$\sigma_{X}=\sqrt{\sigma^{2}_{X}}=\sqrt{\lambda}=\sqrt{6}=2.4495\approx2.45$

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