Answer:

Step-by-step explanation:
Given:
°
From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.

Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,

From the Δ BCD, which is a right angled triangle,

Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.
Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles
and EF is a tangent to the circle at point C.
Therefore, 
Again, using the same theorem as above,

Hence, all the angles are as follows:

Answer:
4
Step-by-step explanation:
Mean = sum of scores /number of scores
We have four scores given ,so, the sum of the scores given will be divided by the number of scores which is 4
Answer:
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:
And let
represent the sample mean, the distribution for the sample mean is given by:
On this case
Solution to the problem
We want this probability:
The best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
Situation satisfies the criteria for the use of hypergeometric distribution. Since no replacement is made, binomial distribution is not applicable (probability does not remain constant).
A=number of target wattage bulbs
B=number of non-targeted wattage bulbs
a=number of target wattage bulbs selected
b=number of non-targeted wattage bulbs selected
P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)
where C(n,x)=combination of x items chosen from n=n!/(x!(n-x)!)
For all following problems,
A+B=4+5+6=15
a+b=3 (selected)
(a) Target wattage = 75W
A=6, B=9, a=2, b=1
P(a,b,A,B)
=P(2,1,6,9)
=C(6,2)*C(9,1)/C(15,3)
=15*9/455
=27/91
(b) target wattage = each of the three
Probability = sum of probabilities of choosing 3 40,60,75-watt bulbs
P(3x40W)+P(3x60W)+P(3x75W)
Case (A,B,a,b)
3x40W (4,11,3,0)
3x60W (5,10,3,0)
3x75W(6,9,3,0)
P(3x40W)+P(3x60W)+P(3x75W)
=C(4,3)*C(11,0)/C(15,3)+C(5,3)*C(10,0)/C(15,3)+C(6,3)*C(9,0)/C(15,3)
=4*1/455+10*1/455+20*1/455
=34/455
Can also be solved by elementary counting, for example, for (a),
P(2x75W)
=C(3,2)*6/15*5/14*9/13
=(3)*6/15*5/14*9/13
=27/91 as before