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If you deposit $30 at the end of every month for 30 years(65 - 35), or 30 x 12 =360 months at 4.5% compounded monthly, then you should have =$22,781.58 in your IRA when you retire at 65.

$22,781.58 - [$30 x 360] = $11,981.58 - Interest earned on your IRA

GuestOct 14, 2019

-1

sorry incorrect answer

GuestOct 14, 2019

You should deposit $453.83 at the end of every 6-month period for 4 years or 8 semi-annual periods.

$453.83 x 8 =$3,630.64 - This is total principal you must deposit.

$4,000 - $3,630.64 = $369.36 - This is the interest earned on the investment.

GuestOct 14, 2019

Since your are given the Financial Formulas to calculate all your posted questions, you should be able to use them and arrive at the right answer. I, however, will use an online financial calculator to give you the answers. You can use the calculator here: https://arachnoid.com/finance/

The monthly deposit will be = $742.82

$3,000,0000 - [$742.82 x 480] = $2,643,446.40 - This is the interest portion of the investment.

GuestOct 14, 2019

+26400

In the diagram below, ADEF is a square with side length 5. If AC = 12
what is BD?

$\text{Let $AF=AD=EF=5$ } \\ \text{Let $AC=12$ } \\ \text{Let $FC=AC-AF= 12-5 = 7$ } $

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{EF}{AC-AF}} &=& \mathbf{\dfrac{AD+BD}{AC}} \\\\ \dfrac{5}{12-5} &=& \dfrac{5+BD}{12} \\\\ \dfrac{5}{7} &=& \dfrac{5+BD}{12} \quad | \quad \cdot 12 \\\\ \dfrac{12\cdot 5}{7} &=& 5+BD \\\\ BD &=& \dfrac{12\cdot 5}{7} -5 \\\\ BD &=& \dfrac{12\cdot 5}{7} -5 \\\\ BD &=& \dfrac{60-5\cdot 7}{7} \\\\ BD &=& \dfrac{60-35}{7} \\\\ BD &=& \dfrac{25}{7} \\ \mathbf{BD} &=& \mathbf{3.5714285714} \\ \hline \end{array}$

$BD \approx 3.57 $

heurekaOct 14, 2019

+118724

The range is the y values, The highest y value (not actually included) is 4. The curled brackets indicates that neither infinity or 4 are actually included.

hene the range is (-infinity, 4)

The arrows are suppose to indicate the direction that the graph will take from that point on.

Hence on the left it goes down for ever and on the right it goes up forever.

Hence the range is (-infinity, +infinity.)

MelodyOct 14, 2019

+2236

Solutions:

a) John has the tiles A, B, E, E, P, T and Z on his tile rack. How many words of seven tiles can he make if he can also make nonsense words like BEEAPTZ ?

Answer: He can make (7!/2!) = 2520 words. The two (2) “E” letters are indistinguishable and the redundancies are factored out by dividing by the factorial(s) of the duplicates.

b) Carlos has the tiles E, F, O, O, R and R on his tile rack. How many words of six tiles can he make if he can also make nonsense words?

Answer: (6!/2!/2!) = 180 words. This is the same algorithm as described above.

c) How many (nonsense) words can Carlos make that begin with a vowel?

Answer: This would depend on the number of tiles, the number of vowels, and the number of duplicates Carlos has on his rack. With the sample space of E, F, O, O, R and R, then starting with E, 5!/2!/2! = 30. Starting with O, 5!/2! =60. The sum of these two sets is the number of words he can make.

d) In the bag there are ten more tiles. They are all different. John's tile rack is empty so he draws five more tiles from the bag. He puts them on the tile rack in the order he drew them. How many different (nonsense) words can John get on his tile rack?

Answer: With all distinguishable tiles, then nPr(10, 5) = 30240 different (nonsense) words.

GingerAleOct 14, 2019

+118724

At a birthday party, 5 boys and 3 girls are seated around a table.

How many different arrangements are possible if no two girls are seated next to each other?

Normally for these questions rotations are considered to be the same.

Reflections are considered to be different

BUT

Are you interested in gender arangements or individual people arrangements?

-------------------------------------

If you are only interested in gender positions then.

Seat a girl first. She is like the begining of the row.

Now there is 5 boys

GBBBBB

We need to slot in 2 more girls

GB*B*B*B*B

The stars indicate where the other 2 girls can go.

4C2 = 6 gender arrangements are possible.

-------------------------------------

If you are interested in the position of individuals then

Stil sit any girl first. (could have chosen a boy if you wanted to)

Now there is 5 boys and 2 girls left to seat.

Sit the 5 boys first. This can be done in 5! = 120 ways

So we have GBBBBB and we have to sit 2 more girls.

GB*B*B*B*B

4C2*2 = 12 ways.

So that is 120*12 = 1440 individual child arrangements.

-----------------------------------------

MelodyOct 14, 2019

+2236

Solution:

$\text {There are $\mathbf{(5-1)!}$ ways to arrange the boys in a circle. } | \small \text {(circular permutation) }\\ \text {There are five $\mathbf{(5)}$ positions between the boys. }\\ \text {There are $\mathbf{P(5,3)}$ ways to arrange the girls to occupy the positions between the boys.}\\ \mathbf{(4!) * P(5,3) = }\\ \mathbf{(24) * (60) = 1440} \\ \text{ }\\ \large \text {There are $\mathbf{1440}$ unique circular arrangements of 5 boys and 3 girls,} \\ \large \text {where each girl is separated from the other girls by at least one boy. }\\ $

GingerAleOct 14, 2019

#1 I think you are right.(The arrow points down towards 5 and going to negative direction, what makes it stop at 4?So I think you are right)

#2 Well the arrows are going into infinitly directions, arrow on the right is raising up towards 10(x-axis) then will contuine forever rising through the y-axisso it is +infinity

The left arrow is going down so it is negative infinity.

GuestOct 14, 2019