A human blinks 6,250,000 times a year.
A seal blinks 1,010 times a year.
Answer:
a. z = 2.00
Step-by-step explanation:
Hello!
The study variable is "Points per game of a high school team"
The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)
The hypothesis is:
H₀: μ ≤ 99
H₁: μ > 99
α: 0.01
There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.
The statistic value under the null hypothesis is:
Z= X[bar] - μ = 101 - 99 = 2
σ/√n 6/√36
I don't have σ, but since this is an approximation I can use the value of S instead.
I hope it helps!
Answer:
6
Step-by-step explanation:
18×(3/6) = 918×(1/6) = 318×(2/6) = 6
Answer: at 11:54
Step-by-step explanation:
Let's define the 10:30 as our t = 0 min.
We know that Train A stops every 12 mins, and Train B stops every 14 mins, they will stop at the same time in the least common multiple of 12 and 14.
To find the least common multiple of two numbers, we must do:
LCM(a,b) = a*b/GCD(a,b)
Where GCD(a, b) is the greatest common divisor of a and b.
In this case the only common divisior of 12 and 14 is 2.
So we have:
LCM(12, 14) = 12*14/2 = 84.
Then the both trains will stop 84 minutes after 10:30
one hour has 60 mins, so we can write 84 minutes as:
1 hour and 24 minutes = 1:24
Then they will stop at the same time at 10:30 + 1:24 = 11:54
