Answer:
$216
Step-by-step explanation:
This question does not involve interest, so all we need to do is to divide the total of payments ($10,368) by the number of payments (48):
$10,368
--------------- = $216/month
48
X and y represent the two students.
Teacher wants to spend at least $5 in each of x and y. That means x or y could be either equal to 5 or higher than 5.
x ≥ 5
y ≥ 5
Teacher only spends under $30. That means the sum of x and y couldn't be equal to higher than 30. It should be lower than 30.
x + y < 30
The correct answer is option D
So, to solve this, we use this equation:
3700 + 0.05(3700)
But, if we want to make it shorter:
1.05(3700)
Now you just multiply(you may want to use a calculator)
1.05(3700) = 3885
She will pay $3885
The answer relies on whether the balls are different or not.
If they are not, which is almost certainly what is intended.
If they are, the perceptive is a bit different. Your
expression gives the likelihood that a particular set of j balls
goes into the last urn and the other n−j balls into the other urns.
But there are (nj) different possible sets of j balls, and each of
them the same probability of being the last insides of the last urn, so the
total probability of completing up with exactly j balls in the last
urn is if the balls are different.
See attached file for the answer.
We have that
<span>tan(theta)sin(theta)+cos(theta)=sec(theta)
</span><span>[sin(theta)/cos(theta)] sin(theta)+cos(theta)=sec(theta)
</span>[sin²<span>(theta)/cos(theta)]+cos(theta)=sec(theta)
</span><span>the next step in this proof
is </span>write cos(theta)=cos²<span>(theta)/cos(theta) to find a common denominator
so
</span>[sin²(theta)/cos(theta)]+[cos²(theta)/cos(theta)]=sec(theta)<span>
</span>{[sin²(theta)+cos²(theta)]/cos(theta)}=sec(theta)<span>
remember that
</span>sin²(theta)+cos²(theta)=1
{[sin²(theta)+cos²(theta)]/cos(theta)}------------> 1/cos(theta)
and
1/cos(theta)=sec(theta)-------------> is ok
the answer is the option <span>B.)
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.</span>
