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ololo11 [35]
3 years ago
14

100% ☺

Mathematics
1 answer:
raketka [301]3 years ago
7 0
Answer :

I believe it’s x = 8 and y = 9

Explanation: hope this helps (got this from somebody else on here)

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Please help !!!!!!!!!
Ainat [17]

Answer:

5

Step-by-step explanation:

3x -12 = -36

-36 ÷ -3 = 12

-7 + 12 = 5

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Sorry, cannot just give you answers without your input.
I'd suggest that you look up online or in your textbook each of the items under "Type of Boundary."  The results you'd get will likely help you fill in the rest of the boxes.
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Neep help as fast as possible. Will mark correct answer brainliest.​
Umnica [9.8K]

Answer:

kbohogohohohohohohohop

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3 years ago
Evaluate 2(x+4)^2 - (3y+5)^3 for x= -3 and y= -1
UNO [17]
<span>2(x+4)^2 - (3y+5)^3
at x=-3 and y=-1

2(-3+4)^2 - ((3*-1)+5)^3
2 (1)^2 - (2)^3
2 - 8
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5 0
3 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
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