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Kobotan [32]
3 years ago
9

-. For each equation below, determine whether

Mathematics
2 answers:
Andre45 [30]3 years ago
7 0

Answer:

Step-by-step explanation:

The first answer a, is not a linear function because of the first term, 2√(x), this is basically 2(x^(1/2)) which means that it is not linear

B is linear because there are no exponents in any of the variables.

C is not linear because it forms a circle when graphed

D is linear because there are no exponents in any of the variables.

E is linear because there are no exponents in angy of the variables.

F is linear because there are no exponents in any of the variables.

G is not linear because it is a quadratic equation.

Arisa [49]3 years ago
6 0

Answer:

b 3x + 4y – 18 = 0  

d y=(x-9)/7

e. x = 25y – 9

f y = 25x

Step-by-step explanation:

First we can eliminate any function with either x or y to a power because that is not linear  (sqrt root is a power)

a  y = 27 sqrt(x)+ 23    Not linear

b 3x + 4y – 18 = 0  

c. x^2 + y^2 = 25   not linear

d y=(x-9)/7

e. x = 25y – 9

f y = 25x

g. y = 0.5x^2 + 10  not linear

b 3x + 4y – 18 = 0    

3x+4y = 18  is in standard form for a linear equation

d y=(x-9)/7

 y = 1/7x - 9/7 is in slope intercept form for a linear equation

e. x = 25y – 9

x -25y = -9   this is in standard form for a linear equation

f y = 25x    this is in slope intercept form for a linear equation

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Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

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t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

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