Answer:yes
Step-by-step explanation:
because 0.89 is a very small number so itd be reasonable for it to be a large number
Answer:The correct answer is A, 14%.
Step-by-step explanation:
All you have to do is take the 30% chance of you HAVING to stop at the first light and find the chance that you won't have to stop at the first light, which is 70%, or .7. Then you take the chance of having to stop at the second light which is 80%, and find the probability of NOT having to stop there, which is 20%, or .2. Then you multiply the probabilities of not having to stop, which equals to .14.
To solve this, we have to find the volume of the cylinder first. The formula to be used is

Given:V= ?r= 6cmh= 10cm
Solution:

V= (3.14)(6cm)

x 10cmV= (3.14)(

) x 10cmV= (

) x 10cmV= 1130.4cm^3
Finding the volume of the cylinder, we can now solve what the weight of the oil is. Using the formula of density, Density = mass/volume, we can derive a formula to get the weight.
Given:Density = 0.857 gm/cm^3Volume = 1130.4 cm^3
Solution:weight = density x volumew= (0.857 gm/cm^3) (1130.4cm^3)w= 968.7528 gm
The weight of the oil is 968.75 gm.
Answer:
15 hours
Step-by-step explanation:
Given:
Fernell's speed
mph
Dabney's speed
mph
Denote:
- distance covered by Fernell
- distance covered by Dabney
- Fernell's time
- Dabney's time
1. If Fernell drove for 3 hours longer than Dabney, then his time is 3 hours more than Dabney's time and

2. If Fernell covered 18 miles less than Dabney, then

Use formula 

Subtract from the first equation the second equation and equate it to 18:


The quotient of r and 12

The quotient of r and 12 as a fraction
