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ss7ja [257]
2 years ago
14

A random sample of 11 university students produced the following data, where x is the minutes spent studying per day, and y is t

he first exam score (out of a maximum of 100 points). The data are presented below in the table of values.x y11 3913 5514 4317 4619 6922 7524 7725 7828 7731 9334 92What is the value of the intercept of the regression line, b, rounded to one decimal place?Provide your answer below:
Mathematics
1 answer:
SIZIF [17.4K]2 years ago
8 0

Answer:

16.373,16.4

Step-by-step explanation:

Given that a  random sample of 11 university students produced the following data, where x is the minutes spent studying per day, and y is the first exam score (out of a maximum of 100 points).

x y

11 39

13 55

14 43

17 46

19 69

22 75

24 77

25 78

28 77

31 93

34 92 Observations 11        

       

ANOVA        

df SS MS F Significance F    

Regression 1 3214.095153 3214.095153 69.46052448 1.59367E-05    

Residual 9 416.4503017 46.27225574      

Total 10 3630.545455      

       

Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%

Intercept 16.37281677 6.48384963 2.525169105 0.0324919 1.705329914 31.04030362 1.705329914 31.04030362

x 2.369323595 0.28428592 8.33429808 1.59367E-05 1.726224167 3.012423023 1.726224167 3.012423023

We get regression line as y =2.369x+16.373

a) value of intercept = 16.373

b) 16.4

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borishaifa [10]

Answer:

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Step-by-step explanation:

Air containing 0.04% carbon dioxide

V, volume of room is 6000 ft3.

Q, rate of air 2000 ft3/min,

initial concentration of 0.4% carbon dioxide,

determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes?

firstly, we find the time taken for air to completely filled the room

Q = V/t

t = V/Q = 6000/2000 = 3min

so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.

there is  an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.

therefore,

3mins = 0.04% of CO2

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1sec = 0.04/180 = 0.00022%/sec

so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec

What is the concentration at 10 minute

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for 0.04% = 6000ft3

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the concentration at 10 minutes= 0.4+0.0133= 0.4133%

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