Any options for that question
∑x = 5 + 7 + 8 + 4 + 11 + 12 + 8 + 7 = 62
(∑x)^2 = 62^2 = 3,844
x bar = 62 / 8 = 7.75
∑x^2 = 25 + 49 + 64 + 16 + 121 + 144 + 64 + 49 = 522
∑y = 79 + 82 + 83 + 81+ 86 + 89 + 91 + 84 = 675
(∑y)^2 = 455,625
y bar = 675 / 8 = 84.375
∑y^2 = 6,241 + 6,724 + 6,889 + 6,561 + 7,396 + 7,921 + 8,281 + 7,056 = 57,069
∑xy = 395 + 574 + 664 + 324 + 946 + 1,068 + 728 + 588 = 5,287
r = (∑xy - n(x bar)(y bar)) / (sqrt(∑x^2 - n(x bar)^2) sqrt(∑y^2 - n(y bar)^2)) = (5,287 - 8(7.75)(84.375)) / (sqrt(522 - 8(7.75)^2) sqrt(57,069 - 8(84.375)^2)) = (5,287 - 5,231.25) / (sqrt(522 - 480.5) sqrt(57,069 - 56,953.125)) = 55.75 / (sqrt(41.5) sqrt(115.875)) = 55.75 / 69.3456 = 0.8039
Answer:
d
Step-by-step explanation:
=1000
10^3 is 1000
Answer:
FV(p)= PV*(1 + g)^t
Step-by-step explanation:
Giving the following information:
Number of insects (PV)= 1,500
Increase rate= 3 weekly
<u>First, we need to calculate the daily growth rate:</u>
Daily rate (g)= [3^(1/7)] - 1
Daily rate (g)= 0.16993
<u>Now, by using the following formula, we can determine the population p in any given day t:</u>
FV(p)= PV*(1 + g)^t
<u>For, example after 7 days:</u>
FV(p)= 1,500*(1.16993^7)
FV(p)= 4,500
<u>For example, after 10 days:</u>
FV(p)= 1,500*(1.16993^10)
FV(p)= 7,206
