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VLD [36.1K]
3 years ago
5

*Due tonight, plz help*

Mathematics
1 answer:
Eduardwww [97]3 years ago
8 0

Answer:

y° = 7°

Step-by-step explanation:

 (8y +4) +(8y +4) +(8y +4)=180°   ( Equilateral Triangle)

24y +12= 180°

24y= 180-12

y=7°

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2. What is the slope of the line through<br> the points (-5, 7) and ({0,-2) ?
valentinak56 [21]

Answer:

-9/5

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-2-7)/(0-(-5))

m=-9/(0+5)

m=-9/5

6 0
2 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

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
Evaluate log12y2, given log12y = 16.
serg [7]

Answer:

Its a i just took the test

Step-by-step explanation:

7 0
3 years ago
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