1. 7 and -7 are both 7 units away from 0. They are 14 units away from each other. Absolute value helps because rather than 7+(-7) being 0 absolute value makes anything positive so it becomes 7+7=14 to be able to count the units between the two.
2. 5 and -5 are both 5 units away from 0. They are 10 units away from each other. Absolute value helps because rather than 5+(-5) being 0 absolute value makes anything positive so it becomes 5+5=10 to be able to count the units between the two.
3. 2 and -2 are both 2 units away from 0. They are 4 units away from each other. Absolute value helps because rather than 2+(-2) being 0 absolute value makes anything positive so it becomes 2+2=4 to be able to count the units between the two.
Hope that helps
10x^2 - 35x....the GCF is 5x
5x(2x - 7) <==
Answer:
12.1
Step-by-step explanation:
The dashed line joining M₁ and M₂ is the hypotenuse of a right triangle as shown in red below.
The base of the triangle is
x₂ - x₁ = 2 - (-3) = 2+ 3 = 5
The height of the triangle is
y₂ - y₁ = 16 - 5 = 11
We can now use Pythagoras' theorem to calculate distance between the two midpoints.
x² = 5² + 11² = 25 + 121 = 146
x = √146 =
12.1
The distance between M₁ and M₂ is 12.1.
Answer:
Step-by-step explanation:
1) Let the random time variable, X = 45min; mean, ∪ = 30min; standard deviation, α = 15min
By comparing P(0 ≤ Z ≤ 30)
P(Z ≤ X - ∪/α) = P(Z ≤ 45 - 30/15) = P( Z ≤ 1)
Using Table
P(0 ≤ Z ≤ 1) = 0.3413
P(Z > 1) = (0.5 - 0.3413) = 0.1537
∴ P(Z > 45) = 0.1537
2) By compering (0 ≤ Z ≤ 15) ( that is 4:15pm)
P(Z ≤ 15 - 30/15) = P(Z ≤ -1)
Using Table
P(-1 ≤ Z ≤ 0) = 0.3413
P(Z < 1) = (0.5 - 0.3413) = 0.1587
∴ P(Z < 15) = 0.1587
3) By comparing P(0 ≤ Z ≤ 60) (that is for 5:00pm)
P(Z ≤ 60 - 30/15) = P(Z ≤ 2)
Using Table
P(0 ≤ Z ≤ 1) = 0.4772
P(Z > 1) = (0.5 - 0.4772) = 0.0228
∴ P(Z > 60) = 0.0228