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Rzqust [24]
3 years ago
10

The sum of three numbers is 10. Two times the second number minus the first number is equal to 12.

Mathematics
2 answers:
eimsori [14]3 years ago
8 0

<u>Answer:</u>

-10,1,19

<u>Step-by-step explanation:</u>

<u></u>

x+y+z = 10 (Equation 1)

2y-x= 12  (Equation 2)

x-y+2z = 7 (Equation 3)

(Equation 2): -x = -2y+12

                       x = 2y-12 (Equation 4)

(Equation 1) - (Equation 3): 2y-z = 3

                                              -z = -2y+3

                                               z = 2y-3 (Equation 5)

Substitute (4) and (5) into (1)

x+y+z = 10

(2y-12)+y+(2y-3) = 10

5y-15 = 10

5y = 5

y=1

Substitute y=1 into (2)

2y-x= 12

2(1)-x= 12

2-x= 12

-x= 12-2

-x= 10

x= -10

Substitute y=1 and x=-10 into (1)

x+y+z = 10

-10+1+z = 10

z-9 = 10

z = 10+9

z = 19

Order: x = -10, y = 1, and z = 19

Vlad [161]3 years ago
4 0

Answer:

-2 5 7

Step-by-step explanation:

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What are the excluded values of x for x+4/-3x2+12x+36
grin007 [14]

Answer:

x = - 2, x = 6

Step-by-step explanation:

The denominator of the rational expression cannot be zero as this would make the expression undefined.

Equating the denominator to zero and solving gives the values that x cannot be.

Solve

- 3x² + 12x + 36 = 0 ( divide through by - 3 )

x² - 4x - 12 = 0 ← in standard form

(x - 6)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

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x = - 2 and x = 6 ← are the excluded values

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3 years ago
Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n
Oksanka [162]

Answer:

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

\hat p=\frac{209}{377}=0.554 estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

p_o=0.5 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

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 true proportion is higher than 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, 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{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

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=0.05. 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>2.097)=0.018  

If we compare the p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

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