If 3 out of 5 are oranges, this means the ratio is 3 oranges for every 5 total.
If there are 12 oranges, this is 4 groups of 3 oranges. You will need 4 groups of 5 total to make it equivalent.
4 x 5 = 20 total fruit.
<u>3</u> x <u>4 </u>= <u>12</u>
5 x 4 20
probability that the card drawn is:
- an ace or a king = 0.1538
- a king or a diamond = 0.3269
<u>Step-by-step explanation:</u>
Here we have , A card is drawn at random from a deck of 52 playing cards. We need to find that Find the probability that the card drawn is:
A deck of standard 52 cards contain four aces. There are four kings in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.
⇒
⇒ 
⇒ 
A deck of standard 52 cards contain four kings. There are 13 Diamonds in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.
⇒
⇒ 
⇒ 
⇒ 
Therefore, probability that the card drawn is:
- an ace or a king = 0.1538
- a king or a diamond = 0.3269
The circumference is about 3.14 (
) times bigger than twice the radius. Mathematically, this can be writted as
, where C equals the circumference and r equals the radius, or
where C equals the circumference and d equals the radius.
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
No it doesn’t satisfy the function
