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muminat
3 years ago
5

Yulian works at the zoo feeding the animals. He puts water in the elephant habitat at the beginning of the day. The table shows

how much water, in gallons, remains at different times during the day. What is the equation that represents Yulian's situation in point-slope form? (HINT: Give all fractions as decimals)

Mathematics
2 answers:
faust18 [17]3 years ago
5 0
<h3>First of all- just to help yall out the verified person is incorrect.</h3>

The answer is-<u>  y= -29= -7/2 (x-6)</u>

<em>i did the test. :)</em>

<h2><u>Please mark me brainiest!</u></h2>

hope i helped

8090 [49]3 years ago
3 0
The slope - intercept form: y = m x + b
43 = 2 m  + b  ⇒   b = 43 - 2 m
36 = 4 m + b 
------------------------
36 = 4 m + 43 - 2 m
36 - 43 = 4 m - 2 m
- 7 = 2 m
m = - 7 : 2
m = - 3.5
b = 43 - 2 · ( - 3.5 ) = 43 + 7 = 50
Answer:
Yulian`s situation is represented by the equation:
y = - 3.5 x + 50
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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).

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3 years ago
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