Answer to Question #243854 in Math for MINU

Question

The temperature of a particular country

n

 is the number of the month, starting at n

=

n=0

 for January. The temperature in the country is greater than 40 o

C

oC

 for months Jun, July, August, and September. It is less than 40 o

C

oC

 for months January, February, March, April, November and December, and 40 o

C

oC

 for months May and October. If a

a

 is a positive integer, then which of the following options might be correct

T

(

n

)

=

a

(

n

−

4

)

(

n

−

9

)

−

40

T(n)=a(n−4)(n−9)−40

 T

(

n

)

=

a

(

4

−

n

)

(

n

−

9

)

−

40

T(n)=a(4−n)(n−9)−40

 T

(

n

)

=

a

(

4

−

n

)

(

9

−

n

)

+

40

T(n)=a(4−n)(9−n)+40

 T

(

n

)

=

a

(

4

−

n

)

(

n

−

9

)

+

40

T(n)=a(4−n)(n−9)+40

 The temperature in December is 40

−

14

a

40−14a

.

 The temperature in April is 40

−

3

a

40−3a

.

Answer 1

$T(0)<40, T(1)<40, T(2)<40, T(3)<40,$

$T(4)=40, T(5)>40, T(6)>40, T(7)>40,$

$T(8)>40, T(9)=40, T(10)<40, T(11)<40.$

a. $T(n)=a(n−4)(n−9)−40$

False

$T(4)=a(4−4)(4−9)−40=-40<0, a>0$

b. $T(n)=a(4-n)(n−9)−40$

False

$T(4)=a(4-4)(4−9)−40=-40<0, a>0$

c. $T(n)=a(4-n)(9-n)+40$

False

$T(4)=a(4-4)(9-4)+40=40, a>0$

$T(9)=a(4-9)(9-9)+40=40, a>0$

$T(0)=a(4-0)(9-0)+40=36a+40>40, a>0$

d. $T(n)=a(4-n)(n−9)+40$

$T(4)=a(4-4)(4-9)+40=40, a>0$

$T(9)=a(4-9)(9-9)+40=40, a>0$

$T(n)<40, a>0,n<4 \ or\ n>9$

$T(n)>40, a>0,4<n<9$

Might be the correct representation

e. The temperature in December is $40−14a.$

$T(11)=40-14a<40, a>0$

Might be the correct representation

f. The temperature in April is $40−3a.$

$T(3)=40-3a<40, a>0$

Might be the correct representation