Answer:
the common ratio is either 2 or -2.
the sum of the first 7 terms is then either 765 or 255
Step-by-step explanation:
a geometric sequence or series of progression (these are the most common names for the same thing) means that every new term of the sequence is created by multiplying the previous term by a constant factor which is called the common ratio.
so,
a1
a2 = a1×f
a3 = a2×f = a1×f²
a4 = a3×f = a1×f³
the problem description here tells us
a3 = 4×a1
and from above we know a3 = a1×f².
so, f² = 4
and therefore the common ratio = f = 2 or -2 (we need to keep that in mind).
again, the problem description tells us
a2 + a4 = 30
a1×f + a1×f³ = 30
for f = 2
a1×2 + a1×2³ = 30
2a1 + 8a1 = 30
10a1 = 30
a1 = 3
for f = -2
a1×-2 + a1×(-2)³ = 30
-10a1 = 30
a1 = -3
the sum of the first n terms of a geometric sequence is
sn = a1×(1 - f^(n+1))/(1-f) for f <>1
so, for f = 2
s7 = 3×(1 - 2⁸)/(1-2) = 3×-255/-1 = 3×255 = 765
for f = -2
s7 = -3×(1 - (-2)⁸)/(1 - -2) = -3×(1-256)/3 = -3×-255/3 =
= -1×-255 = 255
To find the inverse, interchange the variables and solve for y.
f^-1 (x) = 4 + x/2
It’s 10077 because it’s hey
Answer:
Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Step-by-step explanation:
We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.
So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;
Z = 
~ N(0,1)
where, 
= average age of the random sample of horses with colic = 12 yrs
= average age of all horses seen at the veterinary clinic = 10 yrs
= standard deviation of all horses coming to the veterinary clinic = 8 yrs
n = sample of horses = 60
So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( 
12)
P( 12) = P( 
) = P(Z 1.94) = 1 - P(Z < 1.94)
= 1 - 0.97381 = 0.0262
Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
