Answer:
2.25 or 2 1/4 pounds
Step-by-step explanation:
if you add them all up (I converted them to decimals when adding) you get 13.5 or 13 1/2 if your using fractions, then you divide that by 6 because thats how many jars there are and you get 2.25 or 2 1/4
Answer:
The estimated number of times the spinner will land on 3 is 20.
Step-by-step explanation:
The complete question is:
A four-sided spinner is provided. If Tom spins the spinner 80 times then work out an estimate for the number of times the spinner will land on 3.
Solution:
Assume that the four-sided spinner is unbiased.
That is all the four outcomes are equally likely to be selected.
The probability that the spinner lands on any of the four numbers is:
P(1) = P (2) = P (3) = P (4) = 0.25
It is provided that Tom spins the spinner <em>n</em> = 80 times.
The spinner can land on any of the four numbers independently.
The random variable <em>X</em>, defined as the number of times the spinner lands on 3, follows a Binomial distribution with parameters <em>n</em> = 80 and <em>p</em> = 0.25.
The expected value of <em>X</em> is:


Thus, the estimated number of times the spinner will land on 3 is 20.
Let the radius of the circle be r. Then the line from the external point through the center of the circle which extends to the far point on the circle has length 3r .By the tangent - secant theorem
t^2 = 3r * r = 3r^2 ( where t is the length of the tangent).
So t = √(3r^2) = √3r answer.
Answer:
18 since 12 x 2 = 24 so 9 x 2 = 18
Step-by-step explanation:
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)