Answer:
Third quartile (Q₃) = 46.75 minutes.
Therefore, Option (c) is the correct answer.
Step-by-step explanation:
Given: Mean (μ) = 40 minutes and S.D (σ) = 10 minutes
To find : Third quartile (Q₃) = ?
Sol: As the third quartile of normal distribution covers the 75% of the total area of the curve and first quartile covers the 25% of the total area of the curve. Then with the help of z score table, the value represented the third quartile of the normal distribution is:
Q₃ = μ + 0.675 σ
Now by substitution the value of mean and standard deviation,
Q₃ = 40 + 0.675 × (10)
Q₃ = 40 + 6.75
Q₃ = 46.75
Therefore, the third quartile (Q₃) = 46.75. So, option (c) is the correct answer.
Answer:
768 bugs
Step-by-step explanation:
You can rewrite this problem as a function as time where the bug population is f(x), and x is the number of days since the start.
f(x)=6*2^(x/5)
Here, the 6 represents the number of bugs that you start with, the two shows that they double every day, and the /5 shows that they double every 5 days.
By plugging in 35, you get 6*2^7, which is 768.
Answer is B (hope this helps!)
This kinda easy. Every 3 yellow drops is 4 blue drops, so here's the answer. 3 = 4, 6 = 8, 9 = 12, 12 = 16. so you would need 16 blue drops.
