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Thepotemich [5.8K]
3 years ago
9

A person places $26300 in an investment account earning an annual rate of 3.8%, compounded continuously. Using the formula V=Pe∧

{rt}, where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 19 years.
Mathematics
2 answers:
ahrayia [7]3 years ago
6 0

Answer:

54139.76

Step-by-step explanation:

OLEGan [10]3 years ago
5 0

Answer:

V = $ 54135.92

Step-by-step explanation:

The formula for compounding continuously is given as:

V = P(e)^rt

V = Amount of money in the bank after 19 years

P = Principal amount initially invested = $26300

e = base of natural log

r = Interest rate = 3.8% = 0.038

t = time in years = 19

We have to calculate the value of V. Substitute all values into the given formula:

V = P(e)^rt

V = (26300)(e)^(0.038 · 19)

V = (26300)(2.718)^(0.722)

V = (26300)(2.0584)

V = $ 54135.92

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2 years ago
The ratio of teachers to male students to female students in a school is 3:17:18 if the total number of students in the school i
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Answer: 54

Step-by-step explanation:

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8 0
2 years ago
Help me with this question, please!
AURORKA [14]

Answer:

3,432 m²

Step-by-step explanation:

The amount of aluminum in square meters needed to make the mailboxes = 1863(surface area of each mailbox)

Surface area of each mail box = ½(surface area of cylinder) + (Surface area of rectangular prism/box - area of the surface of the box that joins the half-cylinder)

✔️Surface area of ½-cylinder = ½[2πr(h + r)]

r = ½(0.4) = 0.2 m

h = 0.6 m

π = 3.14

Surface area of ½-cylinder = ½[2*3.14*0.2(0.6 + 0.2]

= 0.628(0.8)

Surface area of ½-cylinder = 0.5024 m²

✔️Surface area of the rectangular box/prism = 2(LW + LH + WH)

L = 0.6 m

W = 0.4 m

H = 0.55 m

Surface area = 2(0.6*0.4 + 0.6*0.55 + 0.4*0.55)

Surface area of rectangular box = 1.58 m²

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8 0
3 years ago
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Step-by-step explanation:

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4 0
3 years ago
If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes
Keith_Richards [23]
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
4 0
3 years ago
Read 2 more answers
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