If the beginning temperature was -13°F at 5:00 a. M., what was the temperature at 9:00 a. M.? Change in Temperature 5:00 a. M. – 6:00 a. M. -3°F 6:00 a. M. – 7:00 a. M. -2°F 7:00 a. M. – 8:00 a. M. -6°F 8:00 a. M. – 9:00 a. M. 7°F b. The same cold front hit Hartford, Connecticut the next morning. The temperature dropped by 7°F each hour from 5:00 a. M. To 9:00 a. M. What was the beginning temperature at 5:00 a. M. If the temperature at 9:00 a. M. Was -10°F?

a. The temperature at 9:00 a.m. = -17°F

b. The temperature at 5:00 a.m. = 18°F

Step-by-step explanation:

a. The given information are;

The temperature at 5:00 a.m. = -13°F

The change in temperature from 5:00 a.m. - 6:00 a.m. = -3°F

The change in temperature from 6:00 a.m. - 7:00 a.m. = -2°F

The change in temperature from 7:00 a.m. - 8:00 a.m. = -6°F

The change in temperature from 8:00 a.m. - 9:00 a.m. = 7°F

We have, the temperature at 9:00 a.m. given as follows

At 9:00 a.m., the temperature = -13 - 3 - 2 - 6 + 7 = -17°F

The temperature at 9:00 a.m. = -17°F

b. The temperature drop per hour = -7°F

The temperature at 9:00 a.m. = -10°F

Given that the temperature dropped by 7°F each hour from 5:00 a.m. to 9:00 a.m., which is 4 hours, we have that the temperature at 5:00 a.m. is given by adding 4 × 7°Fto the temperature at 9:00 a.m. as follows;

Therefore, the temperature at 5:00 a.m. = -10 + 7× 4 = 18°F

The temperature at 5:00 a.m. = 18°F.

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58