Answer:Number Sets. There are sets of numbers that are used so often they have special names and symbols: ... The numbers you can make by dividing one integer by another (but not dividing by zero). In other words ... If you square a real number you always get a positive, or zero, result. ... x + 7 = 0, x = −7, Integers, set integer.
Step-by-step explanation:
The series of numbers is adding 5 and then subtracting 2
20+5=25
25-2=23
23+5=28
28-2=26
and so on. Do you understand?
Answer:
The rise is 5
The run is 6
The rate is 5/6
Step-by-step explanation:
Rise is the change in y, or how many units up the line goes. On this graph, y goes up 5. The run is the change in x, or how many units right the line goes; on this graph, it goes 6 units right in between the points. The rate, or slope, is rise over run, so the final answer is 5/6.
The answer to your question is 45 miles / 1 hour * 5280 feet / 1 mile * 1 hour / 3660 seconds =
45 * 5280 * 1 / 1 * 1 * 3660 =
2737600 feet / 3660 second =
64.918
Answer:
Since we can't assume that the distribution of X is the normal then we need to apply the central limit theorem in order to approximate the
with a normal distribution. And we need to check if n>30 since we need a sample size large as possible to assume this.

Based on this rule we can conclude:
a. n = 14 b. n = 19 c. n = 45 d. n = 55 e. n = 110 f. n = 440
Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean
for n=14 or n =19 since the sample size is <30 we don't have enough evidence to conclude that the sample mean is normally distributed
Step-by-step explanation:
For this case we know that for a random variable X we have the following parameters given:

Since we can't assume that the distribution of X is the normal then we need to apply the central limit theorem in order to approximate the
with a normal distribution. And we need to check if n>30 since we need a sample size large as possible to assume this.

Based on this rule we can conclude:
a. n = 14 b. n = 19 c. n = 45 d. n = 55 e. n = 110 f. n = 440
Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean
for n=14 or n =19 since the sample size is <30 we don't have enough evidence to conclude that the sample mean is normally distributed