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nadya68 [22]
3 years ago
11

2x - 3y = 8 -2x + 6y = -20 x = ? y = ? solve the system of equations

Mathematics
1 answer:
Zolol [24]3 years ago
7 0

Answer:

x = 4 + 3/2y

y = -8/3 + 2/3x

1. Finding X

2x-3y=8

2x=8+3y

x+4+3/2y

2. Finding Y

2x-3y=8

-3y=8-2x

y=-8/3+2/3x

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Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

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The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
Question number 8. Prove that 3 + √5 is an irrational number.​
Marina86 [1]

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

\\  \sf \: 3 +  \sqrt{5}  =  \frac{a}{b}   \\  \\   \qquad \: \tiny \sf{(where \:  \: a \:  \: and \:  \: b \:  \: are \:  \: integers \:  \: and \:  \: b \:  \neq \: 0)} \\

\\  \sf \:  \sqrt{5}  =  \frac{a}{b}  - 3 =  \frac{a - 3b}{b}  \\

Since, a, b and 3 are integers. So,

\\ \sf \:  \frac{p - 3b}{b}  \\  \\  \qquad \tiny \sf{ \: (is \:  \: a \:  \: rational \:  \: number \:) } \\

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

\\

Hence, 3 + √5 is an irrational number.

7 0
3 years ago
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