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.
You have to get from $30 to $21.75
21.75/30=0.725 x 100= 72.5% of the original price
Answer:
Step-by-step explanation:
X = number of Yan's marbles. x-3 = number of Geraldo's marbles
annswer: x + (x - 3) = 21
hopes this helps :)
Answer:
45,135,315,225
Step-by-step explanation:
2cos^2x = 1
Divide each side by 2
cos^2x = 1/2
Take the square root of each side
sqrt( cos^2 x) = ±sqrt (1/2)
cos x =±sqrt (1/2)
Make into two separate equations
cos x =sqrt (1/2) cos x = - sqrt(1/2)
Take the inverse cos of each side
cos ^-1 cos (x) = cos ^-1 (sqrt (1/2)) cos ^-1 cos (x) = cos ^-1 (-sqrt (1/2))
x = cos ^-1 (sqrt (1/2)) x = cos ^-1 (-sqrt (1/2))
x = 45 +360 n x = 135+ 360n
x = 315+360 n x =225+360n
Between 0 and 360
45,135,315,225