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2 years ago

Please help me quick !!!hb

Mathematics

1 answer:

Anit [1.1K]2 years ago

4 0

Answer:

Step-by-step explanation:

You basically just factor out the equation using FOIL. You start with the First term 2x and multiply that by 3x. Next, you multiple the 2x with the Outer term which is 5. Then, you multiply the Inner terms 3x and -1 to get -3x. lastly, you use the 5 and -1 and multiply them with each other. The solution would be 6x^2+7x-5 when you combine like terms.

Hope this helps!

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Find the equation of the line. Use exact numbers. y= ( )x + ( )

Rama09 [41]

Answer:

y=-2x+5

Step-by-step explanation:

Use the equation y=mx+b

We can find the b(y-intercept) by looking at the graph

The graph crosses the y-axis at (0,5)

We can look at the slope at the graph. We can see its down 4 over 2 so the slope is -2.

Plug that into the equation y=mx+b

y=-2x+5

6 0

2 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

8 0

3 years ago

4 Points

Vladimir [108]

Answer:

$35$

Step-by-step explanation:

One is given the following function ( $y=3x+5$ ). The given problem asks one to find the output of the function when the input is (10). Keep in mind, a general rule for dealing with functions is that, when present, the variable (x) represents the input. Whereas the variable (y) represents the output. Thus, if one were to substitute (10) in place of (x), and simplify: the result one would get is the output. Apply this idea to the given scenario: