Answer:
0.57142
Step-by-step explanation:
A normal random variable with mean and standard deviation both equal to 10 degrees Celsius. What is the probability that the temperature at a randomly chosen time will be less than or equal to 59 degrees Fahrenheit?
We are told that the Mean and Standard deviation = 10°C
We convert to Fahrenheit
(10°C × 9/5) + 32 = 50°F
Hence, we solve using z score formula
z = (x-μ)/σ, where
x is the raw score = 59 °F
μ is the population mean = 50 °F
σ is the population standard deviation = 50 °F
z = 59 - 50/50
z = 0.18
Probability value from Z-Table:
P(x ≤59) = 0.57142
The probability that the temperature at a randomly chosen time will be less than or equal to 59 degrees Fahrenheit
is 0.57142
Working entirely with the information given in the question,
it's really hard to tell what 'RAW', 'Z', 'T', or '%ile' represent.
Wdym by animal testing? Like doing experiments on rats as an example?
Answer:
y = 4.8
Step-by-step explanation:
y = k/x
4 = k/6
k = 24
y = 24/5
y = 4.8
A = pi * r^2.....r = radius and pi = 3.14
A = (3.14)(13^2)
A = 3.14(169)
A = 530.66
