Answer:
The 2 represents that each toppings costs $2.
Answer: C) x = 2
2^{2x + 2} = 2^{3x}
Since both terms (above) have the same base, set the exponents to be equal:
2x + 2 = 3x (Rearrange to solve for x)
x = 2
∴ x = 2
Answer:
it should be 8 and 5, the first one
Answer:
The probability is 0.683
Step-by-step explanation:
To calculate this, we shall be needing to calculate the z-scores of both temperatures
mathematically;
z-score = (x-mean)/SD
From the question mean = 78 and SD = 5
For 73
z-score = (73-78)/5 = -5/5 = -1
For 83
z-score = (83-78)/5 = 5/5 = 1
So the probability we want to calculate is within the following range of z-scores;
P(-1 <z <1 )
Mathematically, this is same as ;
P(z<1) - P(z<-1)
Using the normal distribution table;
P(-1<z<1) = 0.68269 which is approximately 0.683
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
