Answer:
p = -5
every element in the sequence is created by subtracting 3 from the pervious element.
Step-by-step explanation:
I am not sure you put every necessary information here.
but I'd the visible information is truly everything, then it's is trivial.
the difference between 4 and 1 is ... well, -3. meaning we subtract 3 from 4 to get 1.
we suspect this is the rule and keep trying.
1 -3 = -2
hey it works.
and -8 -3 = -11
hey, also correct.
and the difference between -2 and -8 is -6, and when we place another item in between (p), we cut that in half again to -3. so, it is all consistent.
therefore,
p = -2 -3 = -5
the rule is
an = an-1 - 3
Answer:
cost = 1.35n
$33.75 for 25 songs
Step-by-step explanation:
Based on the numbers given, the cost is proportional to the number of songs downloaded. The constant of proportionality is the cost of one song: 1.35.
cost = 1.35n
For 25 songs, ...
cost = 1.35·25 = 33.75 . . . . dollars
__
The equation is ...
cost = 1.35n
<em></em>Factors of 52 = 1, 2, 4, 13, 26, 52.
The greatest factor of 52 that is a prime number is 13.
Answer:
4 dozens
Step-by-step explanation:
<em>Find Eleanor and Joanna's ratio of eggs to milk</em>
Eleanor:
<em>Dozen eggs : gallons of milk</em>
9 : 3
3 : 1
Joanna:
<em>Dozen eggs : gallons of milk</em>
2 : 2
1 : 1
<em>They gather one gallon of milk each which means Joanna has 1 dozen eggs and Eleanor has 3 dozens of eggs.</em>
Total dozen of eggs = 3 + 1 = 4
Therefore, total dozen of eggs the family have per week is 4 dozens.
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:
And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
