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
Answer: Statement p is false.
Step-by-step explanation:
In both cases, we need to isolate the variables:
p: -3*x + 8*x - 5*x = x
(-3*x - 5*x) + 8*x = x
-8x + 8*x = x
0 = x
This will be true only for one value of x, so this is not always true, which means that the statement is false.
q: (3*x)*(5*y) = 15*x*y
let's solve the left side:
3*x*5*y = 15*x*y
(3*5)*(x*y) = 15*x*y
15*x*y = 15*x*y
This is true for every value of x and y, then this statement is true.
Step-by-step answer:
A line with slope (or gradient) m, passing through a point (x0,y0) can be represented by the equation:
(y-y0) = m(x-x0) ......................(1)
Here,
(x0,y0) = (1,7), and
slope/gradient = -2
Substitute values into (1) above gives
(y-7)=2(x-1)
Expand and simplify
y-7 = 2x-2
y =2x -2 + 7
y=2x+5
is the final answer.
Adjacent angles :
IGJ and IGH
KGL and LGM
vertical angles :
JGK and HGM
IGJ and MGL
hope this helps
