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Diano4ka-milaya [45]
3 years ago
6

(1,-2),(2,1),(3,6),(5,22) a relation or function

Mathematics
2 answers:
tia_tia [17]3 years ago
7 0

Answer:

function

Step-by-step explanation:

It's a function because the input values {1, 2, 3, 5} are all different.

GrogVix [38]3 years ago
3 0

Answer:

A function

Step-by-step explanation:

This relation is a function because every x input has exactly one y output

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Answer:

C

Step-by-step explanation:

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If m(-12) =0 is the result of M(x) using the reminder theorem then what do -12 and 0 represent?
lions [1.4K]

Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that

Solving each of these equations for x we get x = -a0/a1 and x = -b0/b1 respectively, so in order for both equations to be satisfied simultaneously we must have a0/a1 = b0/b1, which can also be written as a0b1 - a1b0 = 0. Formally we can regard this system as two linear equations in the two quantities x0 and x1, and write them in matrix form as

Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.

Now consider two polynomials of degree 2. In this case we seek a single value of x such that

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3 0
3 years ago
In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use
seropon [69]

Answer:

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

p_v =P(Z>7.36)=9.19x10^{-14}  

Since the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .  

Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

\hat p=\frac{492}{778}=0.632 estimated proportion of people that chose a window seat.

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the majority of adults prefer window seats when they fly:  

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(Z>7.36)=9.19x10^{-14}  

5) Conclusion

Since the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .  

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3 years ago
Mike's Music Soul 287 CDs on the first day of a two day sale the store so 96 more CD's on the second day then the first day how
almond37 [142]
The answer is 670.
this is because 96 plus 287 is 383.
so he sold 287 on the 1st day, and 383 on the second day.
383 plus 287 equals 670.
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