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guajiro [1.7K]
3 years ago
7

Consider the line y=4x-4.

Mathematics
1 answer:
erica [24]3 years ago
4 0

9514 1404 393

Answer:

  • parallel: y = 4x -6
  • perpendicular: y = -1/4x +27/4

Step-by-step explanation:

If we want the new line to be written in slope-intercept form, we need to find the new value of the y-intercept. The equation of the line is ...

  y = mx +b . . . . . . . for slope m and y-intercept b

Solving for b gives ...

  b = y -mx . . . . . . . subtract mx from both sides.

The values of x and y come from the point we want the line to pass through. The value of m will be the same for the parallel line as for the given line: 4. For the perpendicular line, it will be the opposite reciprocal of this: -1/4.

<u>Parallel line</u>

  b = 6 -4(3) = 6 -12 = -6

  y = 4x -6

Perpendicular line

  b = 6 -(-1/4)(3) = 6 +3/4 = 27/4

  y = -1/4x +27/4

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You work it backwards.

-- If there were 4 children and each child got 4 pieces,
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-- That's what was left after he took 2 pieces for himself. 
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7 0
3 years ago
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A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
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Answer:

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

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

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

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
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53.58 is 47 percent of what?
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Answer:114

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Step-by-step explanation:

8 0
3 years ago
Y = 6x - 4; x = 8 find the value of y for the given value of x
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X=8
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