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3 years ago

Mia invested $52,000 in an account paying an interest rate of 3.1% compounded

Mathematics

1 answer:

kirill [66]3 years ago

8 0

Answer:

Step-by-step explanation:

Maia will have 3.1 percentage because we divide three 52,000 by 3.1 percent

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Step-by-step explanation:

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3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago

According to a study done by a university student, the probability a randomly selected individual will not cover his or her mou

Sergio [31]

Using the binomial distribution, it is found that:

a) There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when sneezing.

b) There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when sneezing.

c) 9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when sneezing.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

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.

The values of the parameters are given as follows:

n = 18, p = 0.267.

Item a:

The probability is P(X = 6), hence:

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

$P(X = 6) = C_{18,6}.(0.267)^{6}.(0.733)^{12} = 0.1618$

There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when sneezing.

Item b:

The probability is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).

Then:

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

$P(X = 0) = C_{18,0}.(0.267)^{0}.(0.733)^{18} = 0.0037$

$P(X = 1) = C_{18,1}.(0.267)^{1}.(0.733)^{17} = 0.0245$

$P(X = 2) = C_{18,2}.(0.267)^{2}.(0.733)^{16} = 0.0758$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0037 + 0.0245 + 0.0758 = 0.104.

There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when sneezing.

item c:

We have to look at the mean and the standard deviation, given, respectively, by:

E(X) = np = 18 x 0.267 = 4.81.

$\sqrt{V(X)} = \sqrt{18(0.267)(0.733)} = 1.88$

9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when sneezing.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

2 years ago