Answer:
1,333 frogs
Step-by-step explanation:
okay so we know that mass remains conserved no matter where you are.
same here ^^
if the total mass of brass is 200 kg
the total mass of the new frogs formed from it when put together will be the same :)
and if there were n such frogs formed
we have,
total mass = n × mass of each frog
200 = n × 3/ 20
n = 4000/ 3
so it becomes something like 1,333.33
that is nearly 1,333 frogs can be made out of 200 kg of brass each weighing 3/ 20kg (the last ones a bit less to make upto 200kg)
Answer:
0.337 = 33.7% probability that one of the eight children has a food allergy.
Step-by-step explanation:
For each children, there are only two possible outcomes. Either they have a food allergy, or they do not. The probability of a child having food allergy is independent of any other child. This means that we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
7% of U.S. children 4 years of age or younger have a food allergy.
This means that 
A day care program has capacity for 8 children in that age range.
This means that 
What is the probability that one of the eight children has a food allergy?
This is P(X = 1).
0.337 = 33.7% probability that one of the eight children has a food allergy.
Since the number ranges between 3 and 10, an equation can't be made because the number is always unknown or can be different every time
Answer:
x=0
Step-by-step explanation:
I must assume that your goal is to "solve for x." State your goal explicitly each time you post a problem.
Combining like terms, we get 4x = 0, so that x = 0.
Check: Is 5(0) - 13 = 0 - 13 true? YES
Answer:
10 losses
Step-by-step explanation:
Here, we want to get the greatest possible number of games the team lost
Let the number of games won be x
Number drawn be y
Number lost be z
Mathematically;
x + y + z = 38
Let’s now work with the points
3(x) + 1(y) + z(0) = 80
3x + y = 80
So we have two equations here;
x + y + z = 80
3x + y = 80
The greatest possible number of games lost will minimize both the number of games won and the number of games drawn
We can have the following possible combinations of draws and wins;
26-2
25-5
24-8
23-11
22-14
21-17
21-17 is the highest possible to give a loss of zero
Subtracting each sum from 38, we have the following loses:
10, 8, 6, 4, 2 and 0
This shows the greatest possible number of games lost is 10
