I want to be challenged more, so could anyone give me some 4th grade math problems
2 answers:
You might be interested in
Given:
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).
If we draw the diagonals of the octagonal gazebo, the 4 diagonals divide the octagon into 8 triangles.
Note that each triangle is an isosceles triangle whose equal sides are x, the radius of the circle.
The top angle of each triangle is obtained by dividing the full angle by 8.
So, each top angle =
= 45°
Now, in fig., consider one of the triangles Δ OAB. Draw an altitude OC from O to the opposite side AB.
This altitude OC bisects the top angle 45°.
Therefore, ∠ AOC = 22.5°.
Now, in Δ AOC,
So, AC = x sin 22.5°
Note that, AB = 2 AC.
Therefore, AB = 2x sin 22.5°.
Also,
So, OC = x cos 22.5°.
Area of Δ AOB =
= × (2x sin 22.5°) × (x cos 22.5°)
= (2 sin 22.5° cos 22.5°)
= sin 45°
= /
Area of the octagonal gazebo = 8 × one triangular area
= 8 × ( /
)
Area required for mulch = circular area - area of the gazebo
Now, cost per unit area = $1.50.
Hence, total cost g(m) = area × cost per unit area
Total cost g(m) = × 1.5
Hence, total cost g(m) = .
