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ratelena [41]
2 years ago
14

Ceasar opens a bank account and makes an initial deposit of $800. The banker tells Ceasar that he is going to receive an annual

rate of 10% compounded continuously on his investment. Find the bank balance assuming Ceasar leaves the account untouched for 8 years.
Can you please do the problem using the formula A(t)=pe^(rt).
Mathematics
1 answer:
Vikki [24]2 years ago
6 0
Ceasar opens a bank account and makes an initial deposit of $800. The banker tells Ceasar that he is going to receive an annual rate of 10% compounded continuously on his investment. Find the bank balance assuming Ceasar leaves the account untouched for 8 years.


Can you please do the problem using the formula A(t)=pe^(rt).
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Work out the area of this circle.
Troyanec [42]

Answer:

36π mm²

Step-by-step explanation:

Formula: πr²

r=radius

r=6

π6²=36π

6 0
2 years ago
Plz help I am giving away 20 points!!!!!!!!
vova2212 [387]

Okay, so on number one I have no idea. I keep getting 20 because 14 - (-3)2 = 14- (-6) which is really 14 + 6=20.

Number 2 is r+15=61

Number 3 I didn't get any of those answers because you have to solve what is in the parenthesis first so3*5= 15+1=16 then you multiply that by two giving you 16*2=32 and 42-32=10

6 0
3 years ago
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X+8&gt;18<br><br>Solve the inequality
tigry1 [53]

Answer:

Alright well the Answer to your question is

x > 10 Hope this helps :)

Step-by-step explanation:


7 0
3 years ago
ONLY ANSWER IF YOU KNOW THE ANSWER JUST FOR POINTS. Thank you! :)
Ksivusya [100]
Since triangle DEF = triangle JKL, m<D = m<J, m<E = m<K, m<F = m<L.
m<F = m<L = 90 degrees
m<K = m<E = 5(m<D)
but m<E + m<D = 90 degrees        [right angled triangle]
5(m<D) + m<D = 90 degrees
6(m<D) = 90 degrees
m<D = 90 / 6 = 15 degrees.
5 0
3 years ago
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All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0
3 years ago
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