Answer:
The answer is below
Step-by-step explanation:
a) Let x represent the time taken to drive to see the relatives and let d be the distance travelled to go, hence:
60 mi/h = d/x
d = 60x
When returning, they still travelled a distance d, since the return trip takes 1 h longer than the trip there, therefore:
40 mi/h = d/(x+1)
d = 40(x + 1) = 40x + 40
Equating both equations:
60x = 40x + 40
60x - 40x = 40
20x = 40
x = 40/20
x = 2 h
The time taken to drive there = x = 2 hours
b) The time taken for return trip = x + 1 = 2 + 1 = 3 hours
c) The distance d = 60x = 60(2) = 120 miles
The total distance to and fro = 2d = 2(120) = 240 miles
The total time to and fro = 2 h + 3 h = 5 h
Average speed = total distance / total time = 240 miles / 5 h = 48 mi/h
A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2 and 5.1.
Answer: The correct option is B) about 34%
Proof:
We have to find 
To find , we need to use z score formula:
When x = 4.2, we have:
When x = 5.1, we have:
Therefore, we have to find 
Using the standard normal table, we have:
= 
or 34.13%
= 34% approximately
Therefore, the percent of data between 4.2 and 5.1 is about 34%
Answer:
c) 300.9 ft²
Step-by-step explanation:
He has 4 1/4 bags of mulch in his shed to use. Each bag of mulch will cover 70.8 ft².
Hence:
1 bag of mulch = 70.8ft²
4 1/4 bags of mulch = x ft²
Cross Multiply
1 bag × x ft² = 4 1/4 × 70.8
x ft² = 4 1/4 × 70.8/1
x = 300.9 ft²
Option c is correct.
