Answer:
12 miles
Step-by-step explanation:
Let the amount of hours it takes Leroy to catch up to Kevin equal x. Since speed multiplied by time is distance, Leroy would have run a total of 6x miles(6*x). Kevin already had a head start of 24 minutes, which is the same as 2/5 of an hour. In 2/5 an hour, Kevin ran 2 miles as 2 is 2/5 *5. For the next x hours his distance would be a total of 5x(5*x). This means, after x hours, Kevins distance is 5x + 2. In this case, catching up to someone is the same as having the same distance value. This means we can set the two expressions equal to each other, and solve:
6x = 5x + 2
x = 2
The time it took Leroy to catch up is 2 hours, but we are looking for the distance, which is represented by 6x. We substitute in 2 to get:
6(2) = 12
This means Leroy had to run 12 miles.
X = 2. 11 to the 4th power (2(2)) is equal to 14641.
Answer:
72.1 cm
Step-by-step explanation:
The arc length formula is s = rФ, where Ф is the central angle and is in radians (not degrees).
Here we convert 140° to radians: 140°(π/180°) = 2.44 radians.
Then the desired arc length is s = rФ = (29.5 cm)(2.44 radians) = 72.1 cm
Answer:
a) 90 stamps
b) 108 stamps
c) 333 stamps
Step-by-step explanation:
Whenever you have ratios, just treat them like you would a fraction! For example, a ratio of 1:2 can also look like 1/2!
In this context, you have a ratio of 1:1.5 that represents the ratio of Canadian stamps to stamps from the rest of the world. You can set up two fractions and set them equal to each other in order to solve for the unknown number of Canadian stamps. 1/1.5 is representative of Canada/rest of world. So is x/135, because you are solving for the actual number of Canadian stamps and you already know how many stamps you have from the rest of the world. Set 1/1.5 equal to x/135, and solve for x by cross multiplying. You'll end up with 90.
Solve using the same method for the US! This will look like 1.2/1.5 = x/135. Solve for x, and get 108!
Now, simply add all your stamps together: 90 + 108 + 135. This gets you a total of 333 stamps!
Using a binomial distribution considering there's a 30% chance it will rain on any of the three days:
<span>The probability of it raining on 0 days is (1)(0.7)(0.7)(0.7) = 34.3%. </span>
<span>The probability of it raining on 1 day is (3)(0.3)(0.7)(0.7) = 44.1%. </span>
<span>The probability of it raining on 2 days is (3)(0.3)(0.3)(0.7) = 18.9%. </span>
<span>The probability of it raining on 3 days is (1)(0.3)(0.3)(0.3) = 2.7%. </span>
<span>There's a 65.7% chance that it will rain at least once over the three-day period.</span>
