A linear function has no restriction on range, unless there is one on the domain.
The range is <em>all real numbers</em>.
Answer: 16 litres
Step-by-step explanation:
A: 1l x km
B: 1l x+5 km
Running 400 km A used 400/x litres
Running 400 km B used 400/(x+5) litres
400/x- 400/(x+5)= 4
400*(x+5)- x*400-4*x*(x+5)=0
400*x+2000-400*x-4*x²-20*x=0
2000-4*x²-20*x=0 difide by 4 both sides of equation
500-x²-5*x=0
Lets solve the equation using discriminant:
D=5²-(-1)*4*500=2025
sqrt(D)=45
x1= (5+45)/(-2)= -25 x2=(5-45)/(-2)=20
x1=-25 x2=20
x1<0 so is not the solution of the problem ( number of litres can't be negative)
So A uses litr per x=20 km and B uses 1 litr per 20+5=25 km
Running 400 km B uses 400/25=16 litres
Answer:
C. 14
Step-by-step explanation:
F(x) = ∫₀²ˣ √(t³−15) dt
Use second fundamental theorem of calculus:
F'(x) = √((2x)³−15) d/dx (2x)
F'(x) = 2 √(8x³−15)
Evaluate at x=2:
F'(2) = 2 √(8×2³−15)
F'(2) = 2 √(64−15)
F'(2) = 2 √49
F'(2) = 14
Answer:
I believe for the second one it's B, then for the third one it's D
Step-by-step explanation:
Oh also be careful w ur lunch # showing
To answer this
problem, we use the binomial distribution formula for probability:
P (x) = [n!
/ (n-x)! x!] p^x q^(n-x)
Where,
n = the
total number of test questions = 10
<span>x = the
total number of test questions to pass = >6</span>
p =
probability of success = 0.5
q =
probability of failure = 0.5
Given the
formula, let us calculate for the probabilities that the student will get at
least 6 correct questions by guessing.
P (6) = [10!
/ (4)! 6!] (0.5)^6 0.5^(4) = 0.205078
P (7) = [10!
/ (3)! 7!] (0.5)^7 0.5^(3) = 0.117188
P (8) = [10!
/ (2)! 8!] (0.5)^8 0.5^(2) = 0.043945
P (9) = [10!
/ (1)! 9!] (0.5)^9 0.5^(1) = 0.009766
P (10) = [10!
/ (0)! 10!] (0.5)^10 0.5^(0) = 0.000977
Total
Probability = 0.376953 = 0.38 = 38%
<span>There is a
38% chance the student will pass.</span>
