Answer: C
Step-by-step explanation:
im guessing, so if wrong please let me know
Answer:
0.09
Step-by-step explanation:
Given :
P(bike) = 0.8
P(car) = 0.2
P(Late given car) = P(Late | car) = 0.05
P(Late given bike) = p(Late | bike) = 0.1
Probability that professor is late :
P(late) = [P(Late | car) * p(car)] + [p(Late | bike) * p(bike)]
P(late) = [0.05 * 0.2] + [0.1 * 0.8]
P(late) = 0.01 + 0.08
P(late) = 0.09
Answer:
y=19/3
Step-by-step explanation:
<h3>What is the greatest common factor of 15x³y² and 20x⁴y⁴? </h3>
<em>First we have to find the factors of 15x³y²</em>
<em>15x³y² = 3 * 5x³y² </em>
<em>Then we have to find the factors of 20x⁴y⁴</em>
<em>20x⁴y⁴ = 2²x¹y² * 5x³y²</em>
<em>Now we have to find the common factors to both numbers.</em>
<em>The common factors are </em><em>5x³y²</em>
Answer : GCF = (15x³y²,20x⁴y⁴) = 5x³y²
Hope this helps!
Answer:
The speed of the first train is 45 mph and the speed of the second train is 75 mph
Step-by-step explanation:
Let x represent the speed of the first train in mph. Since the second train, is 30 mph faster then the first, therefore the speed of the second train is (x + 30).
The first train leaves at 1:00 pm, therefore at 6:00 pm, the time taken is 5 hours. Therefore the distance covered by the first train at 6:00 pm = x mph * 5 hours = 5x miles
The second train leaves at 3:00 pm, therefore at 6:00 pm, the time taken is 3 hours. Therefore the distance covered by the second train at 6:00 pm = (x + 30) mph * 3 hours = (3x + 90) miles
Since the second train overtakes the first at 6:00 pm, hence:
3x + 90 = 5x
2x = 90
x = 45
Therefore the speed of the first train is 45 mph and the speed of the second train is 75 mph (45 mph + 30 mph).