What are the roots of the equation?

Equation of the line contains the points (-3,2) and (5,-5)

makkiz [27]

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

$(x_ {1}, y_ {1}): (- 3,2)\\(x_ {2}, y_ {2}) :( 5, -5)$

So the slope is:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-5-2} {5 - (- 3)} = \frac {-7} {5 + 3} = - \frac {7} {8}$

Thus, the equation of the line is of the form:

$y = - \frac {7} {8} x + b$

We substitute one of the points and find "b":

$2 = - \frac {7} {8} (- 3) + b\\2 = \frac {21} {8} + b\\b = 2- \frac {21} {8}\\b = \frac {16-21} {8}\\b = - \frac {5} {8}$

Finally, the equation is:

$y = - \frac {7} {8} x- \frac {5} {8}$

Answer:

$y = - \frac {7} {8} x- \frac {5} {8}$

6 0

3 years ago

Solve the following equation 2x=12

Luden [163]

Let's solve your equation step-by-step.
2
x
=
12
Step 1: Divide both sides by 2.
2
x
2
=
12
2
x
=
6
Answer:
x
=
6

6 0

1 year ago

Please help and thank you

Scilla [17]

The answer is B.

Hope this helps!

6 0

3 years ago

According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se

belka [17]

Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

50% of adult smokers started smoking before 21 years old.

This means that $p = 0.5$

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that $n = 5$ .

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875$

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

$P(X \leq 4) = 1 - P(X = 5)$

In which

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

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875$

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

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

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

4 0

2 years ago

Can you please help me?

Nimfa-mama [501]

A=iii
B=iv
C=i
D=ii

That's it. I think you can figure out why this is the case.

5 0

3 years ago