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3 years ago

Find the equation of the line passes through (-3,-4) and whose slope is 3/2 in slope intercept form. (Algebra 2 or Geometry)

Mathematics

1 answer:

ioda3 years ago

4 0

Hey there! I'm happy to help!

Slope intercept form is y=mx+b, where m is the slope and b is the y-intercept. We already have the slope, so our equation so far is y=3/2x+b. We just need to find the b for it to be complete.

To do this, we plug in a point on the line and solve for b. We have a point (-3,-4), so let's use it and solve.

-4=3/2(-3)+b

Undo parentheses.

-4=-9/2+b

Add 9/2 to both sides

b=1/2

Therefore, our equation is y=3/2x+1/2.

Have a wonderful day! :D

You might be interested in

Determine if the following system of equations has no solutions, infinitely many

DochEvi [55]

Answer:

There is one solution

Step-by-step explanation:

2x + y = -1

8x + 3y = -2

Multiply the first equation by -4

-8x -4y = 4

Then add the equations together to eliminate x

-8x -4y = 4

8x + 3y = -2

--------------------

-y = 2

Multiply by -1

y =-2

Now find x

2x+y =-1

2x+-2 =-1

Add 2 to each side

2x-2+2=-1+2

2x=1

Divide by 2

x = 1/2

8 0

2 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$