Answer:
The total cost of 1 belt and 1 tie is $13 and $39 respectively.
Step-by-step explanation:
Given that,
The cost of 4 belts and 5 ties is $247.
Each tie costs 3 times as much as a belt.
Let the cost of a belt is x and that of a tie is y.
ATQ,
4x + 5y = 247 ...(1)
y = 3x ....(2)
Put the value of y from equation (2) in equation (1)
4x + 5(3x) = 247
4x + 15x = 247
19x = 247
x = 13
Put the value in equation (2)
y = 3x
= 3(13)
= 39
So, the total cost of 1 belt and 1 tie is $13 and $39 respectively.
Answer:
z^2 = -12y
or y = -(z^2/12)
Step-by-step explanation:
Answer:
z^2 = -12y
Step-by-step explanation:
Given the vertex at origin;
This is same as (0,0)
The focus (0,-3)
The general form is;
(z-h)^2 = 4p(y-k)
(h,k) is the vertex (0,0); so h = 0 and k = 0
The focus is given as;
(h,k + p)
so ;
k + p = -3
but k =0
so p = -3
Thus, the equation of the parabola is;
(z-0)^2 = 4(-3)(y-0)
z^2 = -12y
Answer:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) = (0.9938)^10
Step-by-step explanation:
Since the annual rainfall is normally distributed,
Given: that
Mean (µ )= 40
and σ = 4.
Let X be normal random variables of the annual rainfall.
P(that there will be over 10 years or more before a year with a rainfall above 50 inches)
P(>50) = 1-P[X ≤50]
1 - P[X- μ/σ ≤ 50 - 40/4]
=1 - P [Z≤ 5/2]
=1 -Φ(5/2)
=1 - 0.9939
= 0.0062
P( the non occurrence of rainfall above 50 inches)
= 1-0.0062
=0.9938
ASSUMPTION:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) =
Answer:
i dont know the answer off the top of my head but im "answering" this because the more people answer a question it will pop up at the top for people to see
Step-by-step explanation:
