1. 60,30,90 right triangle. y will be hypotenuse/2, x will be
hypotenuse*sqrt(3)/2. So x = 16*sqrt(3)/2 = 8*sqrt(3), approximately 13.85640646
y = 16/2 = 8
2. 45,45,90 right triangle (2 legs are equal length and you have a right angle).
X and Y will be the same length and that will be hypotenuse * sqrt(2)/2. So
x = y = 8*sqrt(2) * sqrt(2)/2 = 8*2/2 = 8
3. Just a right triangle with both legs of known length. Use the Pythagorean theorem
x = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13
4. Another right triangle with 1 leg and the hypotenuse known. Pythagorean theorem again.
y = sqrt(1000^2 - 600^2) = sqrt(1000000 - 360000) = sqrt(640000) = 800 5. A 45,45,90 right triangle. One leg known. The other leg will have the same length as the known leg and the hypotenuse can be discovered with the Pythagorean theorem. x = 6. y = sqrt(6^2 + 6^2) = sqrt(36+36) = sqrt(72) = sqrt(2 * 36) = 6*sqrt(2), approximately 8.485281374
6. Another 45,45,90 triangle with the hypotenuse known. Both unknown legs will have the same length. And Pythagorean theorem will be helpful.
x = y.
12^2 = x^2 + y^2
12^2 = x^2 + x^2
12^2 = 2x^2
144 = 2x^2
72 = x^2
sqrt(72) = x
6*sqrt(2) = x
x is approximately 8.485281374
7. A 30,60,90 right triangle with the short leg known. The hypotenuse will be twice the length of the short leg and the remaining leg can be determined using the Pythagorean theorem.
y = 11*2 = 22.
x = sqrt(22^2 - 11^2) = sqrt(484 - 121) = sqrt(363) = sqrt(121 * 3) = 11*sqrt(3). Approximately 19.05255888
8. A 30,60,90 right triangle with long leg known. Can either have fact that in that triangle, the legs have the ratio of 1:sqrt(3):2, or you can use the Pythagorean theorem. In this case, I'll use the 1:2 ratio between the unknown leg and the hypotenuse along with the Pythagorean theorem.
x = 2y
y^2 = x^2 - (22.5*sqrt(3))^2
y^2 = (2y)^2 - (22.5*sqrt(3))^2
y^2 = 4y^2 - 1518.75
-3y^2 = - 1518.75
y^2 = 506.25 = 2025/4
y = sqrt(2025/4) = sqrt(2025)/sqrt(4) = 45/2
Therefore:
y = 22.5
x = 2*y = 2*22.5 = 45
9. Just a generic right triangle with 2 known legs. Use the Pythagorean theorem.
x = sqrt(16^2 + 30^2) = sqrt(256 + 900) = sqrt(1156) = 34
10. Another right triangle, another use of the Pythagorean theorem.
x = sqrt(50^2 - 14^2) = sqrt(2500 - 196) = sqrt(2304) = 48
The number of silver and bronze medals that France won are 9 silver and bronze medals.
Since there are 60 medals awarded, and there were 20 each of bronze, silver, and gold. This means that there are 20 gold, 20 bronze and 20 silver.
Since U.S. swim team won 13 gold medals and Spain won 5gold medals, then the gold won by France will be:
= 20 - (13 + 5)
= 20 - 18
= 2 gold medals
France won 2 gold medals.
Since France won one-third of all the medals, the number of medals won by France will be:
= 1/3 × 60
= 20 medals
Since France won 2 golds medals, then the number of silver and bronze will be:
= 20 - 2
= 18 silver and bronze medals.
Since France won equal number of silver and bronze medals. This will be:
= 18/2
= 9 silver and bronze medals.
Read related link on:
brainly.com/question/19972449
Answer: The table represents a function. What is f(5)?
–8
–1
1
8
Step-by-step explanation:
Answer:
a) 
b) 
c) ![P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589](https://tex.z-dn.net/?f=P%28X%20%5Cleq%2016%29%3D%201-P%28X%3E16%29%20%3D1-P%28X%20%5Cgeq%2017%29%3D%201-%20%5BP%28X%3D17%29%20%2B...%2BP%28X%3D20%29%5D%3D0.589)
d) 
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Part a
For this case we want this probability:
And using the mass function we have this:
Part b
For this case we want this probability:
And we can find this using the complement rule:
And if we replace we got:
Part c
For this case we want this probability:
And we can use the complement rule like this:
![P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%2016%29%3D%201-P%28X%3E16%29%20%3D1-P%28X%20%5Cgeq%2017%29%3D%201-%20%5BP%28X%3D17%29%20%2B...%2BP%28X%3D20%29%5D)
And if we replace we got:
Part d
The expected value is given by:
(x - 4)(x^2 - 5x - 6) =
x(x^2 - 5x - 6) - 4(x^2 - 5x - 6) =
x^3 - 5x^2 - 6x - 4x^2 + 20x + 24 =
x^3 - 9x^2 + 14x + 24 <==
