There is a lack of a diagram to go along with this question...
Answer:
iK ITS A TRICKY QUESTION ITS C THOUGH
Step-by-step explanation:
Answer: the answers are below
Step-by-step explanation:
1 meal = 2/3 of a can of wet dog food, 1/8 of a bag of dry dog food, and 3/5 a patty of special meat
Food for 1 day: 2 meals = 2(2/3) = 4/3 of a can of wet dog food, 2(1/8) = 1/4 of a bag of dry dog food, and 2(3/5) = 6/5 a patty of special meat
The trainer goes to the store and buys 24 cans of wet food, 4 bags of dry food and 3 packages of meat (3*6 = 18 patties)
He has:
there is wet food for: 24/(4/3) = 24*3/4 = 18 days
there is dry food for: 4/(1/4) = 4*4 = 16 days
there is special meat food for: 18/(6/5) = 18*5/6 = 15 days
since the minimum for 15, 16 and 18 is 15 the dog will have food for 15 days and first he will run out of special meat.
Answer:
The mean, because the data distribution is symmetrical.
Step-by-step explanation:
Without even knowing the mean or median this can be solved. The data isn't skewed to the left nor right, so it is symmetrical. The mean is the central value of numbers so it should represent the center well.
Answer: The average was 9 years old find the average age of both groups is 10 years old.
Step-by-step explanation:
Formula foe average : 
Given : The first group of students consists of 10 and their average age was 13 years old.
i.e. 
(1)
The next group consisted of 30 students and their average was 9 years old.
i.e. 
(2)
Then from (1) and (2) , the sum of both groups (first group and next group )students = 130+270 =400
Combined students of both groups (first and next group )= 10+30=40
Now , the average of both groups =
Hence, the average was 9 years old find the average age of both groups is 10 years old.
