You can use systems of equations for this one.
We are going to use 'q' as the number of quarters Rafael had,
and 'n' as the number of nickels Rafael had.
You can write the first equation like this:
3.50=0.05n+0.25q
This says that however many 5 cent nickels he had, and however many
25 cent quarters he had, all added up to value $3.50.
Our second equation is this:
q=n+8
This says that Rafael had 8 more nickels that he had quarters.
We can now use substitution to solve our system.
We can rewrite our first equation from:
3.50=0.05n+0.25q
to:
3.50=0.05n+0.25(n+8)
From here, simply solve using PEMDAS.
3.50=0.05n+0.25(n+8) --Distribute 0.25 to the n and the 8
3.50=0.05n+0.25n+2 --Subtract 2 from both sides
1.50=0.05n+0.25n --Combine like terms
1.50=0.30n --Divide both sides by 0.30
5=n --This is how many NICKELS Rafael has.
We now know how many nickels he has, but the question is asking us
how many quarters he has.
Simply substitute our now-known value of n into either of our previous
equations (3.50=0.05n+0.25q or q=n+8) and solve.
We now know that Rafael had 13 quarters.
To check, just substitute our known values for our variables and solve.
If both sides of our equations are equal, then you know that you have
yourself a correct answer.
Happy math-ing :)
Answer:
The tree is 16.25 m tall.
Step-by-step explanation:
Attached is a diagram that better explains the problem.
From the diagram we see that the distance between the top of the tree and the line of sight of the observer is x.
To find the height of the tree, we need to first find x and then add it to the height of the observers line of sight from the ground.
Using SOHCAHTOA trigonometric function:
tan(20) = x/39.2
=> x = 39.2 * tan(20)
x = 39.2 * 0.364
x = 14.27m
Hence, the height of the tree is:
(14.27 + 1.98)m
16.25m
The tree is 16.25 m tall.
Answer:
(-16.494 ; -3.506)
Step-by-step explanation:
student Prob A Prob B difference, d (A-B)
1 20 35____ - 15
2 30 40 ___ - 10
3 15 20 ___ - 5
4 40 50 __ - 10
Difference, d = -15, -10, -5, -10
Xd = Σd/ n = - 40 / 4 = - 10
Standard deviation of d ; Sd = 4.082
The confidence interval for the difference is given as :
Xd ± Tcritical*(Sd/√n)
Tcritical at 95%; df = n - 1 ; 4 - 1 = 3
Tcritical(0.05, 3)). = 3.182
C.I = -10 ± 3.182(4.082/√4)
C.I = -10 ± 6.494462
C. I = (-16.494 ; -3.506)
