Find the value of x .

20 points! I need the answer ASAP! Thanks!

serious [3.7K]

Answer: y=5/7x+2

Step-by-step explanation:

The y- intercept would be 2 since that is where the line goes through the y axis. Also if you use slope form you will see that the slope is 5/7.

5 0

3 years ago

I need help w/ math due asap plzz

julia-pushkina [17]

Answer:

The answer is D

Step-by-step explanation:

You multiply by 2 19 x 2 = 38 38 x 2 = 76 76 x 2 = 152

then 152 x 2 = D 304

3 0

3 years ago

Order the numbers from least to greatest.<br> 1/8, 5/7, 0.35,0.39, 9/10

Anna11 [10]

Answer:

1/8, 0.35, 0.39, 5/7, 9/10

Step-by-step explanation:

1/8 = 0.125

5/7 = 0.714

0.350

0.390

9/10 = 0.900

now let's order them from least to greatest:

0.125 , 0.350 , 0.390 , 0.714 , 0.900

= 1/8, 0.35, 0.39, 5/7, 9/10

if you have trouble with these types of problems change them all into decimals which will make things easier!

:D

6 0

3 years ago

Read 2 more answers

Please help

katrin2010 [14]

$\dfrac{n^2+3n+2}{n^2+6n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n^2+2n+n+2}{n^2+4n+2n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n(n+2)+1(n+2)}{n(n+4)+2(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{(n+2)(n+1)}{(n+2)(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{n+1}{n+4}-\dfrac{2n}{n+4}=\dfrac{n+1-2n}{n+4}=\dfrac{1-n}{n+4}$

$Answer:\ \boxed{B.\ \dfrac{1-n}{n+4}}$

7 0

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

8 0

3 years ago