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nikitadnepr [17]
3 years ago
10

The graph represents this system of equations. What is the solution to the system of equations represented by the graph? A (0, –

3) B(1, 1) C(1.5, 0 ) D(2.5, –2)

Mathematics
2 answers:
melomori [17]3 years ago
6 0

Answer:

Answer is (2.5,-2)

Step-by-step explanation:

In the graph there are two lines graphed.

We know that linear equations can be solved by graphing the two lines and finding the point of intersection.

The point of intersection of the two lines is where both the equations are satisfied simultaneously.

Hence the solution to the system is the point of intersection.

Here point of intersection of two lines is

x = 2.5 and y = -2

Hence solution is

Option D (2.5,-2)

fomenos3 years ago
5 0
The solution is where the lines intersect....and it is (2.5,-2)
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Q/r - 6 = q for r. <br> Answer is. . . . .
Ksivusya [100]
R = q/q+6


That is going to be the answer
3 0
3 years ago
Which of the following is a solution to the equation 2/3x + 16 = -14?
Gala2k [10]
2/3x+16=-14
        -16  -16
2/3x=-30
  *3    *3
2x=-90
 /2    /2
x=-45

Hope this helped :)

7 0
4 years ago
n a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of inches and a stand
kotykmax [81]

Answer:

(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

(b) The probability that a study participant has a height that is between 67 and 71 inches is 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is 0.0401.

(d) The event in part (c) is an unusual event.

Step-by-step explanation:

<u>The complete question is:</u> In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts​ (a) through​ (d) below. ​(a) Find the probability that a study participant has a height that is less than 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = <u><em>the heights of men in the​ 20-29 age group</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean height = 67.5 inches

            \sigma = standard deviation = 2 inches

So, X ~ Normal(\mu=67.5, \sigma^{2}=2^{2})

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

 

      P(X < 67 inches) = P( \frac{X-\mu}{\sigma} < \frac{67-67.5}{2} ) = P(Z < -0.25) = 1 - P(Z \leq 0.25)

                                                                 = 1 - 0.5987 = <u>0.4013</u>

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

    P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X \leq 67 inches)

    P(X < 71 inches) = P( \frac{X-\mu}{\sigma} < \frac{71-67.5}{2} ) = P(Z < 1.75) = 0.9599

    P(X \leq 67 inches) = P( \frac{X-\mu}{\sigma} \leq \frac{67-67.5}{2} ) = P(Z \leq -0.25) = 1 - P(Z < 0.25)

                                                                = 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = <u>0.5586</u>.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

 

      P(X > 71 inches) = P( \frac{X-\mu}{\sigma} > \frac{71-67.5}{2} ) = P(Z > 1.75) = 1 - P(Z \leq 1.75)

                                                                 = 1 - 0.9599 = <u>0.0401</u>

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

7 0
3 years ago
Dr. Jones has found over the years that 95% of babies delivered weighted x pounds, where |x-8.2| - 1.5 &lt; 0.....Solve the ineq
larisa86 [58]

Answer:

6.7.

Step-by-step explanation:

We have been given an inequality |x-8.2|-1.5 and we are asked to solve our given inequality.

Let us add 1.5 to both sides of our inequality.

|x-8.2|-1.5+1.5

|x-8.2|

Upon using definition of absolute value function |f(x)| we can write our absolute function as:

x-8.2-1.5  

x-1.5+8.2  

x6.7  

Upon combining our ranges we will get,

6.7  

Therefore, our inequality solves to 6.7.



6 0
3 years ago
30.5 $ with a 5% tax rate
Vlad [161]

Answer:

$32.03

Step-by-step explanation:

Since there is a tax rate, you want to multiply $30.50 by 1.05 since you are adding tax to it and it is 5%.  You should get 32.025 as the result, but since you need to round it, it is $32.03.

4 0
3 years ago
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