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3 years ago

50 points for each answer.

Mathematics

2 answers:

AfilCa [17]3 years ago

8 0

Answer:

The correct answer is 12.96

Step-by-step explanation:

You would multiply the compound for every year, given that there is 6 years your answer would be 12.96.

AlekseyPX3 years ago

4 0

Answer:

A=$3.973.42

Step-by-step explanation:

First, convert R as a percent to r as a decimal

r = R/100

r = 2.12/100

r = 0.0212 rate per year,

Then solve the equation for A.

$A = P(1 + r/n)^{nt}$

$A = 3,500.00(1 + 0.0212/4)^{(4)}^{(6)}$

$A = 3,500.00(1 + 0.0053)^{(24)}$

$A = $3,973.42$

Summary:

The total amount accrued, principal plus interest, with compound interest on a principal of $3,500.00 at a rate of 2.12% per year compounded 4 times per year over 6 years is $3,973.42.

Hope this helps! Brainliest would be much appreciated! Have a great day! :)

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3 years ago

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

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

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

$P(X \leq 1) + P(X > 1) = 1$

$P(X > 1) = 1 - P(X \leq 1)$

We also have that:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

3 0

3 years ago