Answer:
Normally distributed => get ready the normal probability table, or equivalent.
mu = mean = 28 oz
sigma = standard deviation = 2 oz
Need to find P(30 to 31), the probability of filling between 30 and 31 oz.
Solution:
With normal probabilities,
P(30 to 31) = P(X<31) - P(X<30), i.e. the difference of the right tails for X=30 and X=31.
Calculate the Z scores
Z(X) = (X-mu)/sigma
Z(31) = (31-28)/2 = 1.5
P(X<31) = P(Z<1.5) = 0.9331928
Z(30) = (30-28)/2 = 1.0
P(X<30) = P(Z<1.0) = 0.8413447
Therefore,
probability of filling between 30 and 31 oz
= P(X<31) - P(X<30)
= 0.9331928 - 0.8413447
= 0.09184805
= 0.0918 (to 4 decimal places)
Step-by-step explanation:
Answer:
where x is the cost of one adult ticket and y is the cost of one child ticket.
Step-by-step explanation:
<em>This is an incomplete question since we would need to know the cost of the adult ticket and the cost of the children ticket.</em>
However, let's say that the price is x dollars per adult and y dollars per child.
Now, we need to find out how much one adult and 6 children paid.
Thus, we would have to multiply the cost per adult by the number of adults and the cost per child per number of children and then sum up these two results.
Writing this in an algebraic way we would have:
Thus, 1 adult and 6 children would have paid x + 6y dollars where x is the cost of the adult ticket and y is the cost of the children ticket.
<em>(For example, if an adult ticket is 6 dollars and a child ticket is 4 dollars we would have that they paid 6 + 6(4) = 6 + 24 = 30 dollars)</em>
Answer:
x=6 y=12
Step-by-step explanation:
30-60-90 triangle
shorter leg is x
longer leg is
hypotenuse is 2x
since the longer leg is than makes x=6
and y=2(6)=12