Solution:
Consider the Given Isosceles Triangle
Considering the Possibilities
Case 1. When two equal angles are of 70°
Let the third angle be x.
Keeping in mind , that sum of Interior angles of Triangle is 180°.
70° + 70° + x= 180°
140° +x= 180°
x= 180°- 140°
x= 40°
Case 2:
When an angle measures 70°, and two equal angles measure x°.
Keeping the same property of triangle in mind, that is sum of interior angles of triangle is 180°.
70° + x° + x° = 180°
⇒ 70° + 2 x° = 180°
⇒ 2 x° = 180° - 70°
⇒ 2 x° = 110°
Dividing both sides by 2, we get
x= 55°
Answer:
the price of adult ticket and student ticket be $10.50 and $8.25 respectively
Step-by-step explanation:
Let us assume the price of adult ticket be x
And, the price of student ticket be y
Now according to the question
2x + 2y = $37.50
x + 3y = $35.25
x = $35.25 - 3y
Put the value of x in the first equation
2($35.25 - 3y) + 2y = $37.50
$70.50 - 6y + 2y = $37.50
-4y = $37.50 - $70.50
-4y = -$33
y = $8.25
Now x = $35.25 - 3($8.25)
= $10.5
Hence, the price of adult ticket and student ticket be $10.50 and $8.25 respectively
<span>measure of ∠EGF = 1/2( 180 - 50)
= 1/2(130)
= 65
</span><span>the measure of ∠CGF = 180 - 65
= 115</span>
3.56 +/- 0.011 = ( 3.549, 3.571). This is the answer.
Answer:
P(A) = 0.2
P(B) = 0.25
P(A&B) = 0.05
P(A|B) = 0.2
P(A|B) = P(A) = 0.2
Step-by-step explanation:
P(A) is the probability that the selected student plays soccer.
Then:
P(B) is the probability that the selected student plays basketball.
Then:
P(A and B) is the probability that the selected student plays soccer and basketball:
P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:
P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.
