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xz_007 [3.2K]
3 years ago
9

What is the result of subtracting the second equation from the first?−4x−2y=-2x−2y=9

Mathematics
1 answer:
FrozenT [24]3 years ago
8 0
The end result would be ( 11/5 , -15/5)
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Solve for b<br><br> -1/3b=9<br><br> B=-27<br> B=-3<br> B=3<br> B=27
JulsSmile [24]

Step-by-step explanation:

-  \frac{1}{3} b = 9 \\ b =  -9 \times 3 \\ b =  - 27

5 0
3 years ago
Graph shown here is the graph of which of the following rational<br> functions?
yarga [219]

Answer:

B) f(x) = 1/x-1

Step-by-step explanation:

You can type it into desmos graphing calculator- hope this helps

6 0
2 years ago
One number is eight less than a second number. Five times the first is 6 more than 6 times the second. Find the numbers.
-Dominant- [34]

Answer:

-42/11

Step-by-step explanation:

x = y - 8

5x = 6 - 6y

So now solve the system of equations, divide everything in the second equation by 5 to get it to x = 6/5 - 6y/5

Now...

x = y - 8

x = 6/5 - 6y/5

Now substitute first equation into the second and x is gonna be -42/11 or the first number

5 0
3 years ago
A road sign has the shape of an equilateral triangle with a side length of 48 inches. What is the height of the sign? The road s
Ivenika [448]

The height of the equilateral triangle is 41. 6 inches. Option C

<h3>How to determine the height</h3>

The formula for finding the height of an equilateral triangle is given as;

h = (a√3)/2

we have a = 48 inches

Let's substitute the value

Height, h = \frac{48\sqrt{3} }{2}

Height = \frac{48 * 1. 732}{2}

Height = \frac{83. 14}{2}

Height = 41. 6 Inches

Thus, the height of the equilateral triangle is 41. 6 inches. Option C

Learn more about equilateral triangles here:

brainly.com/question/1399707

#SPJ1

5 0
1 year ago
For a certain population of men, 8 percent carry a certain genetic trait. For a certain population of women, 0.5 percent carry t
slavikrds [6]

Answer:

C) 150 men and 100 women

D) 200 men and 2000 women

E) 1000 men and 1000 women

Step-by-step explanation:

Hello!

To compare the proportion of people that carry certain genetic trait in men and woman from a certain population two variables of study where determined:

X₁: Number of men that carry the genetic trait.

X₁~Bi(n₁;p₁)

X₂: Number of women that carry the genetic trait.

X₂~Bi(n₂;p₂)

The parameter of interest is the difference between the population proportion of men that carry the genetic trait and the population proportion of women that carry the genetic trait, symbolically: p₁-p₂

To be able to study the difference between the population proportions you have to apply the Central Limit Theorem to approximate the distribution of both sample proportions to normal.

<u><em>Reminder:</em></u>

Be a variable with binomial distribution X~Bi(n;p), if a sample of size n is taken from the population in the study. Then the distribution of the sample proportion tends to the normal distribution with mean p and variance (pq)/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

So for both populations in the study, the sample sizes should be

n₁ ≥ 30

n₂ ≥ 30

Also:

Both samples should be independent and include at least 10 successes and 10 failures.

Both populations should be at least 20 times bigger than the samples. (This last condition is to be assumed because without prior information about the populations is impossible to verify)

  • If everything checks out then (p'₁-p'₂)≈N(p₁-p₂; p(1/n₁+1/n₂))

<u>The options are:</u>

A) 30 men and 30 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 30*0.08= 2.4

Failures: y₁= n₁*q₁= 30*0.92= 27.6

Population 2

Successes: x₂= n₂*p₂= 30*0.5= 15

Failures: y₂= n₂*q₂= 30*0.5= 15

The second condition is not met.

B) 125 men and 20 women

n₁ ≥ 30 but n₂ < 30

Both samples are independent but n₂ is not big enough for the approximation.

C) 150 men and 100 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 150*0.08= 12

Failures: y₁= n₁*q₁= 150*0.92= 138

Population 2

Successes: x₂= n₂*p₂= 100*0.5= 50

Failures: y₂= n₂*q₂= 100*0.5= 50

All conditions are met, an approximation to normal is valid.

D) 200 men and 2000 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 200*0.08= 16

Failures: y₁= n₁*q₁= 200*0.92= 184

Population 2

Successes: x₂= n₂*p₂= 2000*0.5= 1000

Failures: y₂= n₂*q₂= 2000*0.5= 1000

All conditions are met, an approximation to normal is valid.

E) 1000 men and 1000 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 1000*0.08= 80

Failures: y₁= n₁*q₁= 1000*0.92= 920

Population 2

Successes: x₂= n₂*p₂= 1000*0.5= 500

Failures: y₂= n₂*q₂= 1000*0.5= 500

All conditions are met, an approximation to normal is valid.

I hope this helps!

6 0
3 years ago
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