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kari74 [83]
3 years ago
5

Rod is saving Php2000 in a bank at the end of each month which gives an interest of 1% compounded monthly.How much is the saving

s og Rod after 2 years? ​
Mathematics
1 answer:
agasfer [191]3 years ago
7 0

Answer:

Php2040.38

Step-by-step explanation:

Given

P = 2000 --- Principal

r = 1\% --- Rate

t = 2\ years --- Time

n = 12 --- monthly

Required

Determine the amount at the end of two years

This is calculated as:

A = P(1 + \frac{r}{n})^{nt}

So, we have:

A = 2000(1 + \frac{1\%}{12})^{12*2}

A = 2000(1 + \frac{1}{100*12})^{24}

A = 2000(1 + \frac{1}{1200})^{24}

A = 2000(\frac{1200+1}{1200})^{24}

A = 2000(\frac{1201}{1200})^{24}

A = 2000*1.02019

A = 2040.38

<em>Hence, the final amount is: Php2040.38</em>

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mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.96516 probability of a chipmunk living through the year, thus p = 0.96516

Item a:

  • Two is P(X = 2) when n = 2, thus:

P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

  • Six is P(X = 6) when n = 6, then:

P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

  • At least one not living is:

P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

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3 years ago
Mr browner has 7 keys, one for his home 2 for his car and the rest are for his drawers in his office. if he picks up 1 without l
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Answer:

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Step-by-step explanation:

Because there are a total of 7 keys and he has 2 for his car so 2/7.

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Step-by-step explanation:

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3 years ago
The product of 2 √3 and 3 √12 simplified form is...
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2 x 3 x \sqrt{3} x \sqrt{12}

<h3>This means that 2 x 3 is simple, = 6,</h3><h3>But when multiplying \sqrt{3} x \sqrt{12} you simply multply the two numbers and put them in a square root:</h3>

\sqrt{3} x \sqrt{12} = \sqrt{36}

<h3>But this can be square rooted to 6, which then means that the equation is now:</h3>

2 x 3 x 6 = 36

<h3>So the simplified form is 36.</h3>
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X( width): 24/8=3cm

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