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3 years ago

Slope = 4, passing through (-6, 8)

Mathematics

1 answer:

Mademuasel [1]3 years ago

7 0

The point slope form is y - 8 = 4(x + 6) and the slope intercept form would be y = 4x + 32.

In order to find this, we'll start with the base form of point-slope form.

y - y1 = m(x - x1) ----> Plug in the numbers

y - 8 = 4(x - -6) ----> Simplify

y - 8 = 4(x + 6)

Now to find slope intercept form, solve for y.

y - 8 = 4(x + 6) ----> distribute the 4

y - 8 = 4x + 24 -----> add 8 to both sides

y = 4x + 32

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Vanessa throws a tennis ball in the air. the function h(t) = -16t2 + 45t + 7 represents the distance, in feet, that the ball is

antiseptic1488 [7]

Given that height, h(t) of a tennis ball is modeled by the equation h(t)=<span>-16t^2 + 45t + 7, the time taken for the ball to reach maximum height will found as follows:
at maximum height:
h'(t)=0
but from the equation:
h'(t)=-32t+45=0
solving for t we get
t=45/32
t=1.40625~1.4 seconds
Thus the time taken to reach maximum height is 1.4 seconds

</span>

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3 years ago

How to find the probability of exactly one event happening?

Vlad [161]

$probability \: of \: an \: event \: happening = \: number \: of \: ways \: it \: can \: happen \div total \: number \: of \: outcomes$

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3 years ago

Add [ –6 –2 2] + [–3 2 1].

skad [1K]

The answer is... A hope this helps

3 0

3 years ago

Toni paints a wall that is 9 feet tall and 13 feet wide.What is the area of the wall?

aksik [14]

Answer:

117 feet squared

Step-by-step explanation:

Multiply the height (9 feet) by the width (13 feet).

9 x 13 = 117

117 feet squared

Note: squared is added when describing area (it's used as an exponent in numbers)

6 0

2 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

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.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

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

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$

80 miles:

This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

The probability of no damaging accidents is P(X = 0). So

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

6 0

3 years ago