Using the unemployment data​ provided, investigate the association between the male unemployment rate in 2007 and 2010 for a sample of 52 countries. Complete parts a through d. a) Find a regression model predicting the 2010 rate from the 2007 rate for the sample of 52 countries. State in simple language what the model says.
2010 Index=+x 2007 index ​(Round to two decimal places as​ needed.)
State in simple language what the model says. Select the correct choice below and fill in the answer box to complete your choice.​(Round to two decimal places as​ needed.)
The 2010 rate increased by per year.
For each​ 1% increase in the 2010​ rate, the 2007 rate increased by
For each​ 1% increase in the 2007​ rate, the 2010 rate increased by
The 2007 rate increased by per year.
b) Determine the test statistic.
T= ​(Round to two decimal places as​ needed.)
Determine the​ P-value.
P=​(Round to three decimal places as​needed.)
Make a conclusion.
Since the​ P-value is (1) the significance​ level, α​, (2) the null hypothesis. The association (3) significant.
​c) What percentage of the variability in the 2010 Index is accounted for by the regression​ model?
The regression model accounts for nothing ​% of the variability in the 2010 Index.
​(Round to one decimal place as​ needed.)
Country Male 2007Male 2010
1 24.122.4
2 11.810.7
3 8.711.7
4 19.617.7
5 26.822.2
6 13.913.2
7 16.414.3
8 13.510.7
9 12.111.8
10 30.529.5
11 9.513.3
12 22.218.8
13 8.58.2
14 14.911.3
15 22.815.1
16 28.416.6
17 23.84.2
18 25.326.4
19 13.826.1
20 18.618.8
21 8.34.3
22 15.220.8
23 8.88.2
24 19.18.2
25 21.316.8
26 11.29.7
27 11.312.3
28 15.211.5
29 22.216.3
30 21.820.9
31 6.86.2
32 9.98.5
33 12.312.1
34 22.817.2
35 19.721.1
36 18.115.3
37 38.335.4
38 12.114.3
39 22.6 24.8
40 22.1 18.5
41 22.1 21.6
42 34.6 30.1
43 12.8 10.4
44 19.7 17.2
45 22.7 20.8
46 16.3 15.7
47 7.7 7.6
48 5.1 5.7
49 65.3 63.8
50 19.8 19.7
51 11.812.4
52 12.312.8

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

correct answer is c. just did it

step-by-step explanation: