The answer for number 4 is 13 and number 6 is 121 and the last problem for number 7 is 112
Answer: the tuition in 2020 is $502300
Step-by-step explanation:
The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.
The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = $20500
The fee in 2018 is the 19th term of the sequence. Therefore,
T19 = $45,4120
n = 19
Therefore,
454120 = 20500 + (19 - 1) d
454120 - 20500 = 19d
18d = 433620
d = 24090
Therefore, an
equation that can be used to find the tuition y for x years after 2000 is
y = 20500 + 24090(x - 1)
Therefore, at 2020,
n = 21
y = 20500 + 24090(21 - 1)
y = 20500 + 481800
y = $502300
I think in commutative properties of multiplication:
70 x 6000 = 70 x 6000
420,000 = 420,000
I don't know associative properties, only commutative properties. Sorry :(
Any value which is more than 2 standard deviations away from the mean is considered to be "unusual."
2 standard deviations above the mean 52.4 mp would be 52.4+2(1.8), or 56; 2 std devs below the mean would be 52.4 - 2(1.8), or 48.8. Thus, any value larger than 56 or any value smaller than 48.8 would be "unusual."
54.8, 49.1 and 51.3 are not unusual; 56.5 is unusual, because it's greaster than 56.
