Answer:
14 month
Step-by-step explanation:
Lets create equations for these two gyms. I believe this is an algebra 1 problem.
Community Gym: y=70x+50
Workout Gym: y=60x+190
Because both gyms are equal to y, we can set them together and solve for x.
70x+50=60x+190
10x=140
x=14
You can find the cost by plugging in 14 into both of the equations. You get a cost of $1030 after 14 months
Therefore, after 14 months you will have paid the same amount for both gyms.
Answer:
b
Step-by-step explanation:
the answer is b
in the problem you can put the expression into a calculator and you will get the answer 3
in option b, y+2=3 bc a variable is always equal to one so your basically adding 1+2 which is equal to 3
I HOPE I HELPED A BIT<3
Answer:
6 pieces of candy
Step-by-step explanation:
Since an equal number of pieces of candy were given to children on Halloween, and obviously, a different amount of boys and girls were present on Halloween. Since 30 pieces of candy were given out to boys and 48 were given out to girls, a certain amount of candy was given to some amount of boys to equal 30, and that same amount of candy was given to another amount of girls to equal 48. Since we are looking for the greatest number of pieces given to each child, we are looking for the greatest common divisor of 30 and 48, which is 6.
Jake's guesses are illustrations of probabilities.
The probability that Jake's guesses are correct is 1/720
The sample size is:
--- 10 different numbers
The probability that Jake's guesses are correct is as follows:
- 1st guess = 1/10
- 2nd guess = 1/9
- 3rd guess = 1/8
So, the required probability is:
Hence, the probability that Jake's guesses are correct is 1/720
Read more about probabilities at:
brainly.com/question/11234923
Answer: A minimum of 5 people will be needed.
Step-by-step explanation:
150/32 = 4.6875
You can't have a partial person, so round up to 5.
It may be possible to have 4 employees working 32 hours and another working 22 hours, but that is not what the question asks exactly. That is still 5 people.
