All categories

3 years ago

Solve graphically the inequality \[ x ≥ 2 \]

1 answer:

6 0

1) Graph the corresponding equation $ x = 2 $; this will split the plane into two regions. One of the region represents the solution set.

2) Select a point situated in any of the two regions obtained and test the inequality. If the point selected is a solution, then all the region is the solution set. If the selected point is not a solution, then the other (second) region represents the solution set.

3) Test: In this example, let us for example select the point with coordinates (3 , 2) which is in the region to the right of the line x = 2. If you substitute x in the inequality $ x ≥ 2 $ by 3 it becomes $ 3 ≥ 2 $ which is a true statement and therefore (3 , 2) is a solution. Hence, we can conclude that the region to the right of the vertical line x = 2 is a solution set including the line itself which is shown as a solid line because of the inequality symbol $ ≥ $ contains the $ = $ symbol. The solution set is represented by the blue hash region in the graph below including the line x = 2.

You might be interested in

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and me

Dovator [93]

Answer:

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

Step-by-step explanation:

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 time interval.

The problem states that:

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.

To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.

There are 5 weekdays, with a mean of 0.1 calls per day.

The weekend is 2 days long, with a mean of 0.2 calls per day.

So:

$\mu = \frac{5(0.1) + 2(0.2)}{7} = 0.1286$

If today is Monday, what is the probability that Ben receives a total of 2 phone calls in a week?

This is $P(X = 2)$ . So:

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

$P(X = 2) = \frac{e^{-0.1286}*0.1286^{2}}{(2)!} = 0.0073$

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

3 0

2 years ago

Help me plz quick only 5 mins

solniwko [45]

Answer:

line B'C' = 1

m<C'D'A' = 90 degrees (it is still a rectangle)

Step-by-step explanation:

Dilation of coordinates:

Original: d:(3,2), a:(3,4), b:(7,4), c:(7,2)

Divide all numbers by 2.

Altered: d:(1.5,1), a:(1.5,2), b:(3.5,2), c:(3.5,1)

8 0

3 years ago

10 pink cards, 10 blue cards, and 10 yellow cards, each color-set labeled 1-10, are placed in a bucket. What is the probability

icang [17]

Answer:

0.00176

Step-by-step explanation:

Probability = 10/30 × 9/29 × 8/28 × 7/27 × 6/26

= 56/31668

8 0

3 years ago

The wheels of a monster truck are 66 inches tall. Find the distance the monster truck travels when the tires make one 360-degree

LuckyWell [14K]

Answer:

<u>207.35 inches</u>

Step-by-step explanation:

The wheel is in the shape of circle. The height, here, basically gives the diameter.

So,

Diameter = 66 in

Radius is half of diameter, so radius is:

66/2 = 33 inches

r = 33

Now, the distance truck travels when 1 rotation is made is the length of the perimeter of the circular wheel, or the circumference.

The formula is:

$C=2\pi r$

Where C is circumference

r is the radius

Substituting, we get the answer as:

$C=2\pi r\\C=2\pi (33)\\C=66\pi\\C=207.35$

The distance covered is about <u>207.35 inches</u>

3 0

2 years ago

Consider a simple example of moral hazard. Suppose that Woodrow goes into a casino to make one bet a day. The casino is very bas

Neko [114]

Answer:

The expected value of the safe bet equal $0

Step-by-step explanation:

$S=\left\{s_1,s_2,...,s_n\right\}$

is a finite numeric sample space and

$P(X=s_k)=p_k$ for k=1, 2,..., n

is its probability distribution, then the expected value of the distribution is defined as

$E(X)=s_1P(X=s_1)+s_2P(X=s_2)+...+s_nP(X=s_n)X)$

What is the expected value of the safe bet?

In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is

S = {100,-100}

Since the coin is supposed to be fair,

P(X=100)=0.5

P(X=-100)=0.5

and the expected value is

E(X) = 100*0.5 - 100*0.5 = 0

6 0

3 years ago