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3 years ago

9|4/3x-2| + 7 = 7 Please help!

Mathematics

1 answer:

Aleksandr [31]3 years ago

3 0

Answer:

Step-by-step explanation:

Given the expression

9|4/3x-2| + 7 = 7

First note that the value in parenthesis can both be negative and positive;

For the positive value, we have;

9(4/3x-2) + 7 = 7

9(4/3x-2) + 7 - 7 = 7 - 7

9(4/3x-2) = 0

Open the bracket

9(4/3x) - 9(2) = 0

36/3 x - 18 = 0

12x = 18

x = 18/12

x = 3/2

For the negative value;

9(-(4/3x-2)) + 7 = 7

9(-4/3x+2) + 7 - 7 = 7 - 7

9(-4/3x+2) = 0

Open the bracket

9(-4/3x) + 9(2) = 0

-36/3 x + 18 = 0

-12x = -18

x = 18/12

x = 3/2

Hence the value of x is 3/2

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<h2>Age Problem</h2><h3>Answer:</h3>

Kassim is $15$ years old this year.

<h3>Step-by-step explanation:</h3>

Let $k$ be Kassim's age this year.

Kassim's age after $30$ years is $k +30$ . Also, $3$ times the age of Kassim is $3k$ . We are told that after $30$ years, Kassim's age will be $3$ times as his age. So, $k +30 = 3k$ .

Solving for $k$ :

$k +30 = 3k \\ k +30 -3k = 3k -3k \\ -2k +30 = 0 \\ -2k +30 -30 = -30 \\ -2k = -30 \\ \frac{-2k}{-2} = \frac{-30}{-2} \\ k = 15$

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3 years ago

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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:

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

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

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Finally

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