Answer:
The probability that a student who is involved in a sports team also participated in the prom dance = 0.1344
Step-by-step explanation:
56% of the students are involved in a sport team
56% = 0.56
According to the question, it is stated that 24% of the students at the school that are involved in a sports team also participated in the prom dance.
24% = 0.24
This means that we are going to find 24% of the original 56%, since 24% of them also participated in the prom dance.
The probability that a student who is involved in a sports team also participated in the prom dance = 0.24 * 0.56
The probability that a student who is involved in a sports team also participated in the prom dance = 0.1344
Answer: The critical value for a two-tailed t-test = 2.056
The critical value for a one-tailed t-test = 1.706
Step-by-step explanation:
Given : Degree of freedom : df= 26
Significance level : 
Using student's t distribution table , the critical value for a two-tailed t-test will be :-

The critical value for a two-tailed t-test = 2.056
Again, Using student's t distribution table , the critical value for a one-tailed t-test will be :-

The critical value for a one-tailed t-test = 1.706
Answer: 8/4
Step-by-step explanation: if I’m wrong just watch the video for a hint it helps
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.
Use elimination and subtitution method to solve the problem.
First, eliminate x and you'll find the value of y
x - 4y = 12
x - y = 0
--------------- - (substract)
-3y = 12
y = 12/-3
y = -4
Second, subtitute -4 as y and you'll find the value of x
x - y = 0
x- (-4) = 0
x + 4 = 0
x = -4
The solution
x,y = -4,-4