That is approximately, in order, a C, a C, a B, a D, an A, a B, and an A.
In grade points that is a 2, a 2, a 3, a 1, a 4, a 3, and a 4.
The average of those numbers is about 2.7, so you have a 2.7.
You can raise that by bringing up the D as it is an outlier here.
You have two triangles, ADC and ABC.
Sides AD and AB are congruent.
Sides DC and BC are congruent.
Side AC is congruent to itself.
By SSS, triangles ADC and ABC are congruent.
Corresponding parts of congruent triangles are congruent.
That means that angles DAC and BAC are congruent.
Angles DCA and BCA are congruent.
Since m<DAC = 32, then m<BAC = 32
Since m<DCA = 41, then m<BCA = 41.
Now you know the measures of two angles of triangle ABC.
The measures of the interior angles of a triangle add to 180.
You can find the measure of angle B.
m<BAC + m<B + m<BCA = 180
32 + m<B + 41 = 180
m<B + 73 = 180
m<B = 107
Answer:
18, 9, 27
Step-by-step explanation:
36 ÷ 2 = 18. Dividing by 2 is the same as multiplying by 50%.
36 ÷ 4 = 9. Dividing by 4 is the same as multiplying by 25%.
Since 25% • 3 = 75%, you can do 9 (from previous) • 3 to get 27.
Answer:
a)120
b)6.67%
Step-by-step explanation:
Given:
No. of digits given= 6
Digits given= 1,2,3,5,8,9
Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be
6P3= 6!/3!
= 6*5*4*3*2*1/3*2*1
=6*5*4
=120
Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:
As the first and last digits can only be even
then the form of number can be
a)2n8 or
b)8n2
where n can be 1,3,5 or 9
4*2=8
so there can be 8 three-digit numbers with both the first digit and the last digit even numbers
And probability = 8/120
= 0.0667
=6.67%
The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !
