Point A is located at (-3, -5)

Shelly and Terrence completed a different number of tasks in a game. Shelly earned 90 points on each task. Terrence's total poin

fgiga [73]

I'm thinking maybe x is supposed to represent the number of tasks completed because it states Shelly earned 90 points on each task. Then when put together it may look like this, let's say there are 4 tasks= Ex.90 times the 4 tasks minus 20 points which would total the points Terrence's has earned. So i believe x is the total tasks. Hope I didn't make it confusing...

6 0

3 years ago

Read 2 more answers

In the graph below, the hyperbola gets close to the red lines but never touches

soldier1979 [14.2K]

Answer:

Option A asymptote.

Step-by-step explanation:

Asymptotes are lines to which the graph approaches very it can come very close to it but will never touch it. Asymptote are the limits of a graph.

IN the given graph of Hyperbola the two blue lines representing the Hyperbola are coming too close to the red lines but not touching it .

The two red lines are called the Asymptotoes of the Hyprbola.It is the limit of the Hyperbola.

Step-by-step explanation:

7 0

2 years ago

Araron picked 14 pints of bottled water into a cooler for the family road trip. How many gallons of water did he pack into the c

Maslowich

Answer:

1.75 gallons

1 pint = 0.125 gallon

With that information, we can write a proportion:

$\frac{1 pint}{0.125 gallons}$ = $\frac{14 pints}{x}$

Cross multiply:

x = 1.75 gallons

4 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:

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