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2 years ago

Solve the system of equations using the method of your choice

Mathematics

1 answer:

Vsevolod [243]2 years ago

5 0

Answer:

(-3, -1); x = -3; y = -1

Step-by-step explanation:

-3x + 5y = 4; x + y = -4

solving via substitution

rearrange second equation: y = -x - 4

-3x + 5(-x - 4) = 4

-3x - 5x - 20 = 4

-8x -20 = 4

-8x = 24

x = -3

- 3 + y = -4

y = -1

(-3, -1)

Check

- 3 + -1 = -4

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V125BC [204]

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so, for the one fraction, you would get 39/10.

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2 years ago

Solve the following system of equations by using the elimination method.

mariarad [96]

Answer:

the answer is (6, 1)

Step-by-step explanation:

x² + y² - 12 x - 2 y + 12 = 0

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2 years ago

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Step-by-step explanation:

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2 years ago

Write and equation of a line perpendicular to y=-5x+8 and passes through (10,4)

Solnce55 [7]

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2 years ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

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2 years ago