Mike discovered that the pool in his backyard is leaking slowly. The pool holds 16,111 gallons of water, and is leaking at a rate of 9 gallons per day. If Mike does not replace the water that has leaked from the pool, how many gallons of water will remain in the pool after 111 days?A.
15,612 gallons
B.
17,110 gallons
C.
16,222 gallons
D.
15,112 gallons
Answers
Answer:
There are 15,112 gallons of water will remaining in the pool ⇒ answer D
Step-by-step explanation:
* Lets explain how to solve the problem
- The rate is the ratio between two related quantities in different units
- The pool is leaking slowly at rate 9 gallons per second that means the
quantity of the water in the pool decreased 9 gallons each day
- The find the remaining gallons in the pool after 111 days, find the
amount of water has leaked from the pool in these days
∵ The rate of leaking is 9 gallons per day
∵ Mike does not replace the water that has leaked from the pool
- The amount of water leaked in some days = the rate of leaking
× the number of days
∵ The number of days is 111 days
∵ The amount of water lacked = 9 × 111 = 999 gallons
∵ The amount of water remaining in the pool after some days =
the total amount in the pool - the amount of leaked in that days
∵ The pool holds 16,111 gallons of water
∴ The amount of water remaining = 16111 - 999 = 15112 gallons
* There are 15,112 gallons of water will remaining in the pool
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Answers
So
Amount of carbon dioxide released by
1 gallon of gasoline = 19.6 pounds
Then
Amount of carbon dioxide released by
14.5 gallons of gasoline = (19.6 * 14.5) pounds
= 284.2 pounds
So an estimated 284.2 pounds of carbon dioxide will be released by 14.5 gallons of gasoline.
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For a field trip, 56 students registered. On the day of the field trip, 7 students were unable to go.
We know that 7/56 could not go therefore 7 ÷ 56 = .125 or round to .13 could not go.
What can she infer about the wingspans of the two types of birds?
Type 1: {18, 24, 20, 22, 26}
Type 2: {24, 21, 19, 26, 30}
A.
Type 1 and Type 2 birds have similar wingspan distributions.
B.
Type 1 and Type 2 birds have somewhat similar wingspan distributions.
C.
Type 1 birds and Type 2 birds do not have similar wingspan distributions.
D.
Type 1 birds and Type 2 birds have identical wingspan distributions.
Answers
Answer:
Step-by-step explanation:
The given data set for type 1 of birds is:
Type 1: {18, 24, 20, 22, 26}
Type 2: {24, 21, 19, 26, 30}
Mean of the type 1 data is:
Data
18 16
24 4
20 4
22 0
26 16
Now, mean average of squares is:
Standard deviation=
Now, the difference of mean and its standard deviation of type 1 data set is:
=22-2.828
Difference =19.172
The given data set for type 2 of birds is:
Type 2: {24, 21, 19, 26, 30}
Mean of the type 2 data is:
Data
24 0
21 9
19 25
26 4
30 36
Now, mean average of squares is:
Standard deviation=
Now, the difference of mean and its standard deviation of type 2 data set is:
=24-3.84
Difference=20.16
Since, the difference of mean and standard deviation of both type 1 and type 2 data set is different, therefore, Type 1 birds and Type 2 birds do not have similar wingspan distributions.
Hence, option C is correct.
Answer:
Type 1 and Type 2 birds have similar wingspan distributions.
Step-by-step explanation:
Answers
10/6 is more than 1 so
2/10<1<10/6
therfor 2/10 is less
the legit way to do it is
convert bottom numbers to same number
number is 30
2/10 times 3/3=6/10
10/6 times 5/5=50/30
6/10<50/30
Answers
c^2 = a^2 + b^2 -2abcos(C)
cos(C) = (a^2 + b^2 - c^2)/2ab
this formula will solve your problem