Answer:
Question 7:
∠L = 124°
∠M = 124°
∠J = 118°
Question 8:
∠Q = 98°
∠T = 98°
∠R = 82°
Question 15:
m∠G = 110°
Question 16:
∠G = 60°
Question 17:
∠G = 80°
Question 18:
∠G = 70°
Step-by-step explanation:
The angles can be solving using Symmetry.
Question 7.
The sum of interior angles in an isosceles trapezoid is 360°, and because it is an isosceles trapezoid
∠K = ∠J = 118°
∠L = ∠M
∠K+∠J+∠L +∠M = 360°
236° + 2 ∠L = 360°
Therefore,
∠L = 124°
∠M = 124°
∠J = 118°
Question 8.
In a similar fashion,
∠Q+∠T+∠S +∠R = 360°
and
∠R = ∠S = 82°
∠Q = ∠T
∠Q+∠T + 164° = 360°
2∠Q + 164° = 360°
2∠Q = 196°
∠Q = ∠T =98°.
Therefore,
∠Q = 98°
∠T = 98°
∠R = 82°
Question 15.
The sum of interior angles of a kite is 360°.
∠E + ∠G + ∠H + ∠F = 360°
Because the kite is symmetrical
∠E = ∠G.
And since all the angles sum to 360°, we have
∠E +∠G + 100° +40° = 360°
2∠E = 140° = 360°
∠E = 110° = ∠G.
Therefore,
m∠G = 110°
Question 16.
The angles
∠E = ∠G,
and since all the interior angles sum to 360°,
∠E + ∠G + ∠F +∠H = 360°
∠E + ∠G + 150 + 90 = 360°
∠E + ∠G = 120 °
∠E = 60° = ∠G
therefore,
∠G = 60°
Question 17.
The shape is a kite; therefore,
∠H = ∠F = 110°
and
∠H + ∠F + ∠E +∠G = 360°
220° + 60° + ∠G = 360°,
therefore,
∠G = 80°
Question 18.
The shape is a kite; therefore,
∠F = ∠H = 90°
and
∠F +∠H + ∠E + ∠G = 360°
180° + 110° + ∠G = 360°
therefore,
∠G = 70°.
Answer:
-2, 1, 2, 3, 4
Step-by-step explanation:
The domain is the x values of the graph.
Let H = Hailey's age and M = Maverick
Hailey is 3 times as old as Maverick: H = 3M
The sum of their ages is 48 years: H + M = 48
Since H = 3M, substitute 3M in place of H in the second equation:
H + M = 48
3M + M = 48
4M = 48
M = 12
Maverick is 12 years old. Hailey = 3M = 3(12) = 36 years old.
Answer: $95450
Step-by-step explanation:
We obviously want to look in the 3rd column.
Income that can be taxed is within interval for 33%, so (0.33)(315000) = $103950.
Subtract the tax credit, and we get $103950 - $8500 = $95450.
Answer:
a) It can be used because np and n(1-p) are both greater than 5.
Step-by-step explanation:
Binomial distribution and approximation to the normal:
The binomial distribution has two parameters:
n, which is the number of trials.
p, which is the probability of a success on a single trial.
If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.
In this question:
So, lets verify the conditions:
np = 201*0.45 = 90.45 > 5
n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5
Since both np and n(1-p) are greater than 5, the approximation can be used.
