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3 years ago

The number of years it will take for $510 to grow to $1,010.18 at 5 percent compounded annually is ____

Mathematics

1 answer:

Alina [70]3 years ago

4 0

Answer:

13.66 years

Step-by-step explanation:

Given data

P=$510

A=$1,010.18

R=5%

t= ln(A/P)/r

susbstitute

t= ln(1,010.18/510)/0.05

t=ln1.980/0.05

t= 0.683/0.05

t= 13.66

Hence the time is 13.66 years

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The difference of two numbers is 15.The larger number is 1 more than 3 times the smaller number. Find the two numbers

lisabon 2012 [21]

Answer:

Step-by-step explanation:

2) let the two numbers be x and y

x-y =15

(taking x to be big)

x = 3 (y)+1

i.e. x-3y =1

Subtracting we get

2y =14 or y =7

x =15+7 =22

Larger number is 22 and smaller number is 7

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

Alysha is 2 years younger than Bryce/

Let Alysha be x years old then Bryce would be x+2

Sum of their ages

=x+x+2 =28

2x+2 =28

2x=26

x=13

Hence Alysha is 13 years old and Bryce 14 years old.

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3 years ago

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Step-by-step explanation:

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A hand-held digital music player was marked down by 1/4 of the original price.

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3 years ago

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

balu736 [363]

Answer:

$P(X \ge 1) = 0.509$

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$