Answer:
Step-by-step explanation:
Given:
A car starts with a dull tank of gas
1/7 of the gas has been used around the city.
With the rest of the gas in the car, the car can travel to and from Ottawa three times.
Question asked:
What fractions of a tank of gas does each complete trip to Ottawa use?
Solution:
Fuel used around the city = 
Remaining fuel after driving around the city = 1 -
= 
According to question:
As from the rest of the gas in the car that is 
, the car can complete 3 trip to Ottawa which means,
By unitary method:
The car can complete 3 trip by using = tank of gas.
The car can complete 1 trip by using = 
=
= 
= tank of gas
Thus, tank of gas used for each complete trip to Ottawa.
Answer:
vertical asymptote is: x=4
horizontal asymptote is: y=1
pls mark as brainliest :)
Answer:
y=3x+5
Step-by-step explanation:
If it’s 9.8 meters then it would be 980 centimeters.
Answer:
a) 0.4121
b) $588
Step-by-step explanation:
Mean μ = $633
Standard deviation σ = $45.
Required:
a. If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
We solve using z score formula
= z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
For x = $646
z = 646 - 633/45
z = 0.22222
Probability value from Z-Table:
P(x<646) = 0.58793
P(x>646) = 1 - P(x<646) = 0.41207
≈ 0.4121
b. How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16? (Round your answer to the nearest dollar.)
Converting 0.16 to percentage = 0.16 × 100% = 16%
The z score of 16%
= -0.994
We are to find x
Using z score formula
z = (x-μ)/σ
-0.994 = x - 633/45
Cross Multiply
-0.994 × 45 = x - 633
-44.73 = x - 633
x = -44.73 + 633
x = $588.27
Approximately to the nearest dollar, the amount should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16
is $588
