Finding an irrational number between which given pair of numbers supports the idea that irrational numbers are dense in real numbers? 3.14 and pi 3.33 and 1/3 e squared and square root of 5 square root of 64 over 2 and square root of 16

So, we need to find irrational numbers between the given pairs.

Remember that the sum between an irrational number and an rational number is irrational.

For example, for the first case, we want a irrational number between:

3.14 and pi:

pi = 3.14159265 is irrational

pi - 0.0001 = 3.14159265... - 0.0001 = 3.14149265...

So this number:

3.14149265...

is an irrational number larger than 3.14 and smaller than pi.

Second cacse:

3.33 and 1/3

(here the range would be actually:

1/3 = 0.33 and 3.33

So we want an irrational number larger tan 0.33 and smaller than 3.33

here we can just use pi = 3.141592...

third case:

e^2 and √5

Firs let's write these numbers so we can see how they look.

e^2  =7.389...

√5 = 2.236

So we want a number larger than 2.236... and smaller than 7.389...

Again, here we can use pi = 3.141592...

2.236... < 3.141592... < 7.389...

final case:

√(64/2) and √16

we have:

√(64/2) = 5.65

√16 = 4

So we want an irrational number larger than 4 and smaller than 5.65

Again, let's use our beloved number pi.

we have that:

pi + 1  is an irrational number:

pi + 1 = 3.14159265... + 1.0 = 4.14159265

This number, 4.14159265..., is irrational, is larger than 4 and is smaller than 5.65, so we found the irrational number between the given pair of numbers.

b). 4320 have a nice day

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.