That would depend on the size of your classroom. but, typically the answer is no
Answer:
600 m³
Step-by-step explanation:
Given:
- length of swimming pool = 30 m
- width of swimming pool = 10 m
- depth at deep end = 3 m
- depth at shallow end = 1 m
To find the volume of the water in the pool, find the area of a cross section (see attached image) and multiply it by the width of the pool.
<u>Area of cross section</u>
= area of rectangle - area of triangle
= (3 × 30) - [1/2 × 30 × (3 - 1)]
= 90 - 30
= 60 m²
<u>Volume of pool</u>
= area of cross section × width
= 60 × 10
= 600 m³
Step-by-step explanation:
answer and explanation is pinned
Solution: We are given:
μ=3.1,σ=0.5,n=50
We have to find P(Mean <2.9)
We need to first find the z score
z= (xbar-μ)/(σ/sqrt(n))
=(2.9-3.1)/(0.5/sqrt(50))
=(-0.2)/0.0707
=-2.83
Now we have to find P(z<-2.83)
Using the standard normal table, we have:
P(z<-2.83)=0.0023
Therefore the probability of the sample mean being less the 2.9 inches is 0.0023