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viktelen [127]
3 years ago
6

Ordered Pairs Please X=2

Mathematics
2 answers:
Sladkaya [172]3 years ago
7 0

(2, 0) (2, 7) assuming that Y can equal anything.

artcher [175]3 years ago
3 0

For this case, we have that by definition, in mathematics, an ordered pair is given by a pair of mathematical objects, in which one element and another is distinguished.

They are usually denoted by:

(x, y)

In this case we want to construct ordered pairs with x = 2.

Then, "y" can take any value of the real numbers to form ordered pairs withx = 2

Examples:

(2, -1); (2.5); (2.80) ...

Answer:

"y" can take any value

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I need help . Plz help me
Gekata [30.6K]
The mean is the same thing as the average. To find the average, add all the numbers up (aka find the sum of the numbers) and divide by how many numbers there is:
average =  \frac{sum \: of  \: numbers}{\# \:  of  \: numbers}

So the sum of your numbers is: -6 + 2 + 5 + -7 + -11 + -6 = -23. And there are 6 numbers total.
That means average =  \frac{-23}{6} = -3.833333 ≈ -3.8

Your final answer is -3.8
5 0
3 years ago
Una maquina que fabrica tornillos produce un 3% de piezas defectuosas. Si hoy se han apartado 51 tornillos defectuosos cuantos t
grin007 [14]

Answer:

1700 tornillos

Step-by-step explanation:

Dado que la máquina produce un 3% de tornillos defectuosos en un día. Produjo 51 tornillos defectuosos en un día.

Deje que el número total de tornillos producidos ese día sea x

Por lo tanto;

3% de x = 51

3/100 * x = 51

x = 51 * 100/3

x = 1700 tornillos

7 0
2 years ago
Does anyone know all the concepts/rules needed to learn math in college? I want to make sure I have things I need now in my seni
Mandarinka [93]

Answer:

Basic college math topics include whole numbers, fractions, numbers, decimals and integers. Problem solving, algebra, percents and geometry are also included in course content.

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
Help Please !!! Math
Keith_Richards [23]
The 3rd one I thinkkkkk ??
6 0
3 years ago
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