For five individuals the blood platelet aggregation was measured before (x1) and after (y1) they performed a certain task. these five pairs of values gave the following results: ¯x1 = 30.2, y¯1 = 39.0, cov(x1, y1) = 88.5. for another six individuals the blood platelet aggregation was also measured before (x2) and after (y2) they performed the same task. these six pairs of values gave the following results: ¯x2 = 52.17, y¯2 = 63.67, cov(x2, y2) = 148.67. find the covariance of the combined 11 pairs.

25.2

step-by-step explanation:

2+4x6-8/10= 25.2

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

true

step-by-step explanation:

60.9774

step-by-step explanation:

because when you add 47.6 plus 7.32 equals 54.92

and when you add .0574 + 6= 6.0574

and when you add 54.92 and 6.0574 you get the answer of 60.9774.

plzz mark brainlest last one got deleted. cause i didn't explain it but i did but i understand what the mod meant