Answer:
a. m<B = 100°
b. m<L = 55°
c. Scale factor = 9/11
Step-by-step explanation:
a) Similar triangles have their three corresponding angles congruent to each other, while the ratio of their corresponding sides are proportional to each other.
Since ∆ABC ~ ∆GLJ, therefore,
<A = <G,
<B = <J
<C = <L
m<J = 100° (given)
Therefore,
m<B = m<J = 100°
m<B = 100°
b) m<A = m<G
m<A = 25° (given)
Therefore,
m<A = m<G = 25°
m<G = 25°
m<L = 180° - (m<J + m<G)
Substitute
m<L = 180° - (100° + 25°)
m<L = 55°
c) scale factor of smaller triangle to the larger = side length of smaller triangle / corresponding side length of bigger triangle
Scale factor = AB/GJ
Substitute
Scale factor = 18/22
Simplify
Scale factor = 9/11
Answer: 0.8238
Step-by-step explanation:
Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with 
and 
.
Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.
Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-
![P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]](https://tex.z-dn.net/?f=P%28x%3E92%29%3D1-P%28x%5Cleq92%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B92-106%7D%7B15%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq%20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq-0.93%29%3D1-%281-P%28z%5Cleq0.93%29%29%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%5Cleq%20-z%29%3D1-P%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%3DP%28z%5Cleq0.93%29%3D0.8238%5C%20%5C%20%5B%5Ctext%7BBy%20using%20z-value%20table.%7D%5D)
Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238
Answer:
a) There is not sufficient evidence to support the claim that the mean attendance is greater than 642.
Step-by-step explanation:
since in the question it is mentioned that the average attendance at games should be more 642 and according to this he moving the team with a larger stadium. Also the hypothesis conducted and the conclusion would be failure to deny the null hypothesis
So here the conclusion that should be made in non-technical term is that there should be no enough proof in order to support the claim that the mean attendence is more than $642
5 and 6 is 65 and 3 and 2 is 25 so I would say add 65 and 25
Answer:
The area is 16.62
Step-by-step explanation:
The expression you use is pi x r squared. which is 3.14 x 2.3 x 2
