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3 years ago

Use the substitution method to solve the system of equations. y = -8x 4x - y = 3 O AG-2 O B. (6-1 oc c. (6.-4 D. G-4

Mathematics

2 answers:

PtichkaEL [24]3 years ago

8 0

Answer:

x = 1/4

y = -2

Step-by-step explanation:

Triss [41]3 years ago

6 0

<h2>Known : </h2>

y = -8x

4x - y = 3

<h2>Asked: </h2>

Use the substitution method to solve the system of equations ?

<h2>Answered:</h2>

<h3>Substitution y</h3>

4x - y = 3

4x - (-8x) = 3

4x + 8x = 3

12x = 3

x = 3/12

x = 1/4

<h3>Substitution x</h3>

4x - y = 3

4(1/4) - y = 3

4/4 - y = 3

1 - y = 3

-y = 3 - 1

-y = 2

y = -2

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lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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3 years ago

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Answer:

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zhannawk [14.2K]

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3 years ago

Use the substitution method to solve the system of equations. y = -8x 4x - y = 3 O AG-2 O B. (6-1 oc c. (6.-4 D. G-4​

Use the substitution method to solve the system of equations. y = -8x 4x - y = 3 O AG-2 O B. (6-1 oc c. (6.-4 D. G-4