ella is making 3 batches of banana milkshake she needs 3/4 gallon of milk for each bath.Her measuring cup holds 1/4 gallon.How m
1 answer:
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Answer:
The diagram for the question is missing, but I found an appropriate diagram fo the question:
Proof:
since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle
∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
∠DOP = 22.5°
∠PDO = 67.5°
∠ADO = 22.5°
∠AOD = 67.5°
Step-by-step explanation:
Given:
AB = CD = 297 mm
AD = BC = 210 mm
BCPO is a square
∴ BC = OP = CP = OB = 210mm
Solving for OC
OCB is a right anlgled triangle
using Pythagoras theorem
(Hypotenuse)² = Sum of square of the other two sides
(OC)² = (OB)² + (BC)²
(OC)² = 210² + 210²
(OC)² = 44100 + 44100
OC = √(88200
OC = 296.98 = 297
OC = 297mm
An isosceless tringle is a triangle that has two equal sides
Therefore for △OCD
CD = OC = 297mm; Hence, △OCD is an isosceless triangle.
The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles
Since BC = OB = 210mm
∠BCO = ∠BOC
since sum of angles in a triangle = 180°
∠BCO + ∠BOC + 90 = 180
(∠BCO + ∠BOC) = 180 - 90
(∠BCO + ∠BOC) = 90°
since ∠BCO = ∠BOC
∴ ∠BCO = ∠BOC = 90/2 = 45
∴ ∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
For ΔOPD
Note that DP = 297 - 210 = 87mm
∠PDO + ∠DOP + 90 = 180
∠PDO + 22.5 + 90 = 180
∠PDO = 180 - 90 - 22.5
∠PDO = 67.5°
∠ADO = 22.5° (alternate to ∠DOP)
∠AOD = 67.5° (Alternate to ∠PDO)
Answer:
from the t-distribution table, at df = 7 and t = 2.23
Lies p-values [ 0.05 and 0.025 ]
Hence;
0.025 < p-value < 0.05
Step-by-step explanation:
Given that;
= 6.5 gpm
μ = 5 gpm
n = eight runs = 8
standard deviation σ = 1.9 gpm
Test statistics;
t = ( - μ) /
we substitute
t = (6.5 - 5) /
t = 1.5 / 0.67175
t = 2.23
the degree of freedom df = n-1 = 8 - 1
df = 7
Now, from the t-distribution table, at df = 7 and t = 2.23
Lies p-values [ 0.05 and 0.025 ]
Hence;
0.025 < p-value < 0.05