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Mila [183]
3 years ago
5

Three rats in a lab are injected a type of medicine. The independent probabilities for each rat exhibiting side effects are 0.5,

0.4 and 0.3 respectively.
Find the probability distribution formula for this problem with random variable X: the number of rats exhibiting side effects (out of 3)​
Mathematics
1 answer:
EleoNora [17]3 years ago
7 0

Step-by-step explanation:

There are four possible values of X: 0 rats show side effects, 1 rat shows side effects, 2 rats show side effects, or all 3 rats show side effects.

Probability X = 0:

P = (1 − 0.5) (1 − 0.4) (1 − 0.3)

P = 0.21

Probability X = 1:

P = (0.5) (1 − 0.4) (1 − 0.3) + (1 − 0.5) (0.4) (1 − 0.3) + (1 − 0.5) (1 − 0.4) (0.3)

P = 0.44

Probability X = 2:

P = (0.5) (0.4) (1 − 0.3) + (0.5) (1 − 0.4) (0.3) + (1 − 0.5) (0.4) (0.3)

P = 0.29

Probability X = 3:

P = (0.5) (0.4) (0.3)

P = 0.06

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X²+2xy+y²=144, (x+y)³?​
Bumek [7]

Answer:

The value of (x + y)³ is 1728.

Step-by-step explanation:

First, you have to find the value of x + y :

x² + 2xy + y² = 144

x² + xy + xy + y² = 144

x(x + y) + y(x + y) = 144

(x + y)(x + y) = 144

(x + y)² = 144

x + y = √144

x + y = 12

Next, the value of (x + y)³ :

(x + y) = 12³ = 1728

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Find the second partial derivatives of the following functions
artcher [175]

(a) <em>z</em> = 3<em>x</em> ² - 4<em>xy</em> + 15<em>y</em> ²

has first-order partial derivatives

∂<em>z</em>/∂<em>x</em> = 6<em>x</em> - 4<em>y</em>

∂<em>z</em>/∂<em>y</em> = -4<em>x</em> + 30<em>y</em>

and thus second-order partial derivatives

∂²<em>z</em>/∂<em>x</em> ² = 6

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = -4

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = -4

∂²<em>z</em>/∂<em>y</em> ² = 30

where ∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = ∂/∂<em>x</em> [∂<em>z</em>/∂<em>y</em>] and ∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = ∂/∂<em>y</em> [∂<em>z</em>/∂<em>x</em>].

(b) <em>z</em> = 4<em>x</em> <em>eʸ</em>

∂<em>z</em>/∂<em>x</em> = 4<em>eʸ</em>

∂<em>z</em>/∂<em>y</em> = 4<em>x</em> <em>eʸ</em>

∂²<em>z</em>/∂<em>x</em> ² = 0

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = 4<em>eʸ</em>

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = 4<em>eʸ</em>

∂²<em>z</em>/∂<em>y</em> ² = 4<em>x</em> <em>eʸ</em>

<em />

(c) <em>z</em> = 6<em>x</em> ln(<em>y</em>)

∂<em>z</em>/∂<em>x</em> = 6 ln(<em>y</em>)

∂<em>z</em>/∂<em>y</em> = 6<em>x</em>/<em>y</em>

∂²<em>z</em>/∂<em>x</em> ² = 0

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = 6/<em>y</em>

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = 6/<em>y</em>

∂²<em>z</em>/∂<em>y</em> ² = -6<em>x</em>/<em>y</em> ²

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3 years ago
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