Answer: C. -25 is the least one, B. 25 is the first one, A. 1/25 is the second, and D. -1/25 is the third. So in a simpler way Least to Greatest is -25, -1/25, 1/25, 25. I hope this helped!

5 0

2 years ago

Read 2 more answers

The population of two different villages is modeled by the equations shown

vlada-n [284]

Answer:

Year = 1995 and population = 315

Year = 2015 and population = 715

Step-by-step explanation:

It is given that the population of two different villages is modeled by the given equations:

$y=x^2-30x+540$

$y=20x+15$

The population of both villages are same after x years after 1980 if

$x^2-30x+540=20x+15$

$x^2-30x+540-20x-15=0$

$x^2-50x+525=0$

Splitting the middle term, we get

$x^2-15x-35x+525=0$

$x(x-15)-35(x-15)=0$

$(x-15)(x-35)=0$

$x=15,35$

It means, after 15 or 35 years, the population will same.

For x=15, years is $1980+15=1995$ and population is

$y=20(15)+15=315$

For x=35, years is $1980+35=2015$ and population is

$y=20(35)+15=715$ .

Therefore, population are equation in Year = 1995 and population = 315 or Year = 2015 and population = 715.

7 0

3 years ago

Try to prove the hypothesis from part B that the sum of a rational and an irrational number is an irrational number.

Maurinko [17]

The next step of your proof is to subtract (a/b) from both sides.

Then you get, x = (m/n) - (a/b)

Since rationals are closed over addition, (m/n) + (-a/b) is a rational number.

Therefore, x (an irrational number) = a rational number <em>This is a false statement which is a contradiction. So, the assumption was incorrect.</em>

Thus, the sum of a rational and irrational number is an irrational number. QED

8 0

3 years ago

A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on th

agasfer [191]

Answer:

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

Step-by-step explanation:

Given that; the probability of each face turning up is proportional to the number of dots on that face

P(1) = 1×P(1)

P(2) = 2×P(1)

P(3) = 3×P(1)

P(4) = 4×P(1)

P(5) = 5×P(1)

P(6) = 6×P(1)

P(T) = 21×P(1)

Where;

P(x) is the probability of getting number x on the dice.

P(T) is the total probability of obtaining any number

N(x) is the number of possible number x in terms of the distribution function.

P(x) = N(x)/N(T) ....1

And since P(T) is constant, and P(T) is proportional to N(T) then,

P(x) is directly proportional to N(x)

So, equation 1 becomes;

P(x) = N(x)/N(T) = P(x)/P(T) ....2

The probability of getting either a 5 or a 2 in one throw

P(2U5) = (P(2) + P(5))/P(T)

Substituting the values of each probability;

P(2U5) = (2P(1) + 5P(1))/21P(1)

P(2U5) = 7P(1)/21P(1)

P(1) cancel out, to give;

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

8 0

3 years ago