a. The events are mutually exclusive, so
b. For the same reason,
Another way to see this is to use the result from part (a), and the inclusion/exclusion principle:
and the right side reduces to 0.
c. Same as (b),
d. Same as (b),
We can also use the distributivity rule for unions and intersections to write
e. If the is the complement of
, then by DeMorgan's law,
We will have that for the previous year, he had in the account:
For the year previous to that one, he had:
For the previous year to that, he had:
For the year previous to that one, he had:
For the second year that the money was in the account, there was:
And the orinial ammount of money that was in the account was:
From this, we know that Jhon had originally $1400.37 in the bank account.
Answer:
15°.
Step-by-step explanation:
1. Angles ADC and CDB are supplementary, thus
m∠ADC+m∠CDB=180°.
Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.
2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then
m∠CDB=m∠CBD=65°.
The sum of the measures of interior angles of triangle is 180°, therefore,
m∠CDB+m∠CBD+m∠BCD=180° and
m∠BCD=180°-65°-65°=50°.
3. Triangle ABC is isosceles, with base BC. Then
m∠ABC=m∠ACB.
From the previous you have that m∠ABC=65° (angle ABC is exactly angle CBD). So
m∠ACB=65°.
4. Angles BCD and DCA together form angle ACB. This gives you
m∠ACB=m∠ACD+m∠BCD,
m∠ACD=65°-50°=15°.
Have a good Day!
