Find h(x)
f(x)+g(x)=5x+15+4x+20=9x+35
ok so if gertrude works 4
f(x)=5x+15
f(4)=5(4)+15
f(4)=35
if we input 4 for x in f(x) to find how much made togethe rin 4 hours
h(4)=9x+35
h(4)=9(4)+35
h(4)=36+35
h(4)=71
split evenly?
71/2=35.5
frankalone=35 for 4 hours
frankwithgertrued=35.5 for 4 hours
slightly bettewr with gerturde
- Make an equation representing the number of vehicles needed.
We have six drivers so
x + y ≤ 6
That's not really an equation; it's an inequality. We want to use all our drivers so we can use the small vans, so
x + y = 6
- Make an equation representing the total number of seats in vehicles for the orchestra members.
s = 25x + 12y
That's how many seats total; it has to be at least 111 so again an inequality,
25x + 12y ≥ 111
We solve it like a system of equations.
x + y = 6
y = 6 - x
111 = 25x + 12y = 25x + 12(6-x)
111 = 25x + 72 - 12x
111 - 72 = 13 x
39 = 13 x
x = 3
Look at that, it worked out exactly. It didn't have to.
y = 6 - x = 3
Answer: 3 buses, 3 vans
1.V = 0.45cm^3
2.V = 0.06cm^3
3.V = 0.39cm^3
4.D = 7.77g/cm^3
Given that:
mean,μ=35.6 min
std deviation,σ=10.3 min
we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.
using z-table, the z-score that will give us 0.2296 is:-1.99
therefore:
z-score is given by:
(x-μ)/σ
hence:
-1.99=(x-35.6)/10.3
-20.497=x-35.6
x=35.6-20.497
x=15.103
Answer: A. x + y = 75 B. She spends 25 minutes running and 50 minutes dancing every day. C. Yes it is.
Step-by-step explanation: B. She spends 25 minutes running and 50 minutes dancing every day, because if she dances for 25 more minutes then she runs then we can minus 25 from 75 which we get 50. Then we take the 50 a divide it by 2 because of the two categories dancing and running. So we know she spends 25 minutes running then 25 minutes dancing but we need to add the 25 minutes to dancing. So she spends 25 minutes running and 50 dancing. C. Yes it is because she already dances for 50 minutes a day so all she needs to do is workout for 10 minutes less so 65 minutes one day.
