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ioda
2 years ago
9

Recall the hurricane relief fund context from the previous investigation. Bob set up a hurricane relief donation fund and starte

d it off by donating $ 35 . Instead of tripling in value each day, suppose it doubled in value each day
Mathematics
1 answer:
Delvig [45]2 years ago
8 0

This question is Incomplete

Complete Question

Recall the hurricane relief fund context from the previous investigation. Bob set up a hurricane relief donation fund and started it off by donating $ 35 . Instead of tripling in value each day, suppose it doubled in value each day

Complete the following table showing the account values at the end of each day.

Number of days since the initial investment, n Amount (in dollars) in the relief account, A

0 1 2 3 4

Answer:

Number of days(n) = Amount (in dollars) in the relief account(A)

0 = $35

1 = $70

2 = $140

3 = $280

4 = $560

Step-by-step explanation:

The formula for doubling time

P(t) = Po(2)^t/k

Where P = Amount after time t

Po = Initial Amount

t = number of days

k = Frequency at which it doubles

For 0 days

t = 0, k = 1, P(o) = $35

P(t) = Po(2)^t/k

P(0) = $35(2)^0/1

= $35(2)⁰

= $35 × 1

= $35

For 1 day

t = 1, k = 1 , P(0) = $35

P(t) = Po(2)^t/k

P(0) = $35(2)^1/1

= $35(2)¹

= $35 × 2

= $70

For 2 days

P(t) = Po(2)^t/k

t = 2, k = 1, P(0) = $35

P(0) = $35(2)^2/1

= $35(2)²

= $35 × 4

= $140

For 3 days

P(t) = Po(2)^t/k

t = 3, k = 1

P(0) = $35(2)^3/1

= $35(2)³

= $35 × 8

= $280

For 4 days

P(t) = Po(2)^t/k

t = 3, k = 1, P(0) = $35

P(0) = $35(2)^5/1

= $35(2)⁴

= $35 × 16

= $560

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Ephemeral services corporation (esco) knows that nine other companies besides esco are bidding for a $900,000 government contrac
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Answer:

c. $100,000

Step-by-step explanation:

Calculation of the expected net profit of Ephemeral services corporation

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4 0
2 years ago
James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo
Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is y = 2 ^n

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒  n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒  for n = 1 , y = 2       ..... (1)

Similarly, if the paper if folded TWICE  ⇒  n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒  for n = 2 , y = 4       ..... (2)

⇒y  = 2^2  =  2^n

Continuing the same way, if the paper is folded SEVEN times  ⇒  n = 7

So, y = 2^ n = 2^7 = 128

⇒  There are total 128 equal parts.

Lastly,  if the paper is folded ELEVEN  times  ⇒  n = 11

So, y = 2^ n = 2^{11} = 2048

⇒  There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is y = 2 ^n

8 0
2 years ago
James is four years younger than Austin. If three times James' age is increased by the square of Austin's age, the result is 28.
lara31 [8.8K]
Let's use J for James's age and A for Austin's age. The equations are:

J = A - 4
3J + A² = 28

Just plug (A - 4) in the place of J in the second equation. This gives you:

3(A - 4) + A² = 28
-->
A² + 3A - 12 = 28
--> 
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-->
A = 5 or -8

-8 is nonsense, so Austin is 5 years old. Therefore, James is 1 year old.
7 0
1 year ago
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