Actually there is enough information to solve this
problem. First, let us find the total per row and per column.
(see attached pic)
P(Grade 10 | opposed) with P(opposed | Grade 10)
P(Grade 10 | opposed) = Number in Grade 10 who are opposed
/ Total number of Opposed (column)
P(Grade 10 | opposed) = 13 / 41 = 0.3171
P(opposed | Grade 10) = Number in Grade 10 who are opposed
/ Total number in Grade 10 (row)
P(opposed | Grade 10) = 13 / 32 = 0.4063
Therefore:
P(Grade 10 | opposed) IS NOT EQUAL P(opposed | Grade 10),
hence they are dependent events.
Answer:
P(Grade 10 | opposed) < P(opposed | Grade 10)
To solve this we use trigonometric functions that would relate the hypotenuse y and the given values. For this case we use cosine function which is expressed as:
cosine theta = adjacent side / hypotenuse
cosine 52 = 35 / y
y = 35 / cos 52
y = 56.85
Answer:
24 cm
Step-by-step explanation:
Given: Picture of the star is made up of 4 unshaded equilateral triangles of equal area and a shaded square.
Area of the shaded region (square) = 
To find: the perimeter of the star
Solution:
Area of the shaded region (square) = 
Area of square = 

As each equilateral triangle has side lengths that are the same as the side length of the square,
length of side of the star = 3 cm
Perimeter is the sum of lengths of the sides.
Perimeter of star = sum of 8 sides of the star
= 8 × 3 = 24 cm
Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes

.
The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.
The probability of selecting one coin is

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.
P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.
Thus

</span><span>P(B) means that the first envelope contains a quarter AND the
second envelope contains a quarter
</span><span>Thus

Therefore,

Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>

</span><span>

</span><span>
Therefore,

</span>
Answer: B.
Step-by-step explanation: TVM Solver Equation:
N = 216 (12 x 18 years)
I% = 3.5
PV = 0
PMT = - $350
FV = 105,106.7593
P / Y = 12 (months)
C / Y = 12
PMT: END