Here are the fuel efficiencies (in mpg) of 12 new cars . 28, 14, 48, 22, 14, 18, 52, 36, 32, 20, 12, 55 What is the percentage o
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25% assuming you don’t count the car with 18 mpg fuel efficiency.
if you add the amount of cars up that have less than 18 mpg and then divide it by the total number (12) you will get a decimal (0.25) to find the percent multiply the decimal by 100
Answer:
a = 2, b = -9, c = 3
Step-by-step explanation:
Replacing x, y values of the points in the equation y = a*x^2 + b*x +c give the following:
(-1,14)
14 = a*(-1)^2 + b*(-1) + c
(2,-7)
-7 = a*2^2 + b*2 + c
(5, 8)
8 = a*5^2 + b*5 + c
Rearranging:
a - b + c = 14
4*a + 2*b + c = -7
25*a + 5*b + c = 8
This is a linear system of equations with 3 equations and 3 unknows. In matrix notation the system is A*x = b whith:
A =
1 -1 1
4 2 1
25 5 1
x =
a
b
c
b =
14
-7
8
Solving A*x = b gives x = Inv(A)*b, where Inv(A) is the inverse matrix of A. From calculation software (I used Excel) you get:
inv(A) =
0.055555556 -0.111111111 0.055555556
-0.388888889 0.444444444 -0.055555556
0.555555556 0.555555556 -0.111111111
inv(A)*b
2
-9
3
So, a = 2, b = -9, c = 3
<span>8x^2-8x-13 is the correct answer </span>
Answer: the value of the account at the end of 6 years is is $8577
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
P = 6000
r = 6% = 6/100 = 0.06
n = 4 because it was compounded 4 times in a year.
t = 6 years
Therefore,.
A = 6000(1+0.06/4)^4 × 6
A = 6000(1+0.015)^24
A = 6000(1.015)^24
A = $8577
Answer:
A ) Not orthogonal to each other
B) 50i + 40j + 105k
C) The tensor product is attached below
D ) The value of X = F.X is attached below
Step-by-step explanation:
attached below is the detailed solution of the above problem
A) for the vectors ( u ) and ( v ) to be orthogonal to each other [ U.V has to be = 0 ] but in this scenario U.V = 4 hence they are not orthogonal to each other
b) The vector normal to plane is gotten by : U x V
= 50i + 40j + 105k