Mean equals average
So 8 plus 10 equals 18
18 divided by 2 is 9
Mean is 9
Answer:
2x² - 12x + 13
Step-by-step explanation:
Given following:
Start solving from right to left across the function.
Steps:
Answer:
(B)
4 +/- 3 sqrt(2) or 4+3sqrt(2) and 4-3sqrt(2)
Step-by-step explanation:
4(c-4)^2=72
4(c-4)(c-4)=72
foil the parenthesis (first, outside, inside, last)
4(c^2 -4c -4c +16)=72
4(c^2-8c+16)=72
divide each side by 4
c^2-8c+16=18
subtract 18 from both sides
c^2-8c-2
use quadratic formula
((-b +/- sqrt((-b^2)-4ac)))/2a
((-(-8)+/-sqrt((-8)^2-4(1)(-2)))/2(1))
(8+/-sqrt(64-(-8)))/2
(8+/-sqrt(64+8))/2
(8+/-sqrt(72))/2
(8+/-sqrt(36 * 2))/2
(8+/-6sqrt(2))/2
4+/-3sqrt(2)
or 4+3sqrt(2) and 4-3sqrt(2)
Answer:
96cm^2
Step-by-step explanation:
We can divide it into two rectangles: 7x8 and 10x(12-8), arriving at the solution of 7cm*8cm + 10cm*4cm = 56cm^2 + 40cm^2 = 96cm^2
We can also divide it into two rectangles: 12x7 + (12-8)x(10-7), arriving at 12cm*7cm + 4cm*3cm = 84cm^2 + 12cm^2 = 96cm^2
We could also divide it to one big rectangle and one "negative" rectangle of 10x12 and -8x(10-7), arriving at 10cm*12cm - 8cm*3cm = 120cm^2 - 24cm^2 = 96cm^2
Answer:


And we can find the limits in order to consider values as significantly low and high like this:


Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
For this case we can consider a value to be significantly low if we have that the z score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is higher tor equal to 2.
For this case we have the mean and the deviation given:


And we can find the limits in order to consider values as significantly low and high like this:

