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3 years ago

How Do Solve 3/4=3/8x-3/2 And Show All Your Work

Mathematics

2 answers:

ipn [44]3 years ago

5 0

The answer is: x = 6

Step 1: Simplify 3/8x to 3x/8. 3/4 = 3x/8 - 3/2

Step 2: Add 3/2 to both sides. 3/4 + 3/2 = 3x/8

Step 3: Simple 3/4 + 3/2 to 9/4. 9/4 = 3x/8

Step 4: Multiply both sides by 8. 9/4 x 8 = 3x

Step 5: Simplify 9/4 x 8 to 72/4. 72/4 = 3x

Step 6: Simplify 72/4 to 18. 18 = 3x

Step 7: Divide both sides by 3. 18/3 = x

Step 7: Simplify 18/3 to get your answer: 6

Let me know if this helps! Have a great rest of your night Jordan!

lubasha [3.4K]3 years ago

3 0

It's too hard for me to show all the work, but you can find a common denominator in 8.

Change 3/4 to 6/8 and -3/2 to -12/8.

Now that all the fractions have the same denominator, you can multiply the whole problem by 8 and cancel out all denominators.

You are left with 6=3x-12

Add 12 to 6 and cancel out on the right side.

18=3x

Divide by 18 by 3

x=6

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ANEK [815]

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The second one

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3 years ago

Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

Formula: $P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}$

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: $P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}$

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

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