A bag contains four red marbles, two green ones, one lavender one, three yellows, and one orange marble. HINT (See Example 7.] H
1 answer:
Answer: 35
Step-by-step explanation:
Given : A bag contains four red marbles, two green ones, one lavender one, three yellows, and one orange marble.
Total = 4+2+1+3+1=11
To find sets of four marbles include none of the red ones, we need to exclude red marbles when we count the total number of marbles.
Then, the total marbles(exclude red) =11-4=7
Now, the combination of 7 marbles taking 4 at a time is given by :-
Hence, the number of sets of four marbles include none of the red ones = 35
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THE QUESTION CAN BE WRITTEN AS THIS;
It is given that ∠ABC and ∠CBD are a____________ So, m∠ABC+m∠CBD=90° using the b____________. It is also given that m∠ABC=35°. Using the substitution property of equality, you have c________+ m∠CBD=90°. Therefore, using the subtraction property of equality, m∠CBD=55°
Answer:
a)COMPLIMENTARY ANGLE
b)DEFINITION OF COMPLEMENTARY ANGLES
c)35°+ m<CBD = 90 degrees.
Step by step Explanation:
a)It's given that <ABC and <CBD are (COMPLIMENTARY ANGLE)
We know that the Sum of Complementary angles equal to 90 degrees.Those angles don't have to be next to each other, just so long as the total is 90 degrees. That's why the Sum of m<ABC and m<CBD equals 90 degrees that is m<ABC + m<CBD = 90 degrees
b)using the (DEFINITION OF COMPLEMENTARY ANGLES) It is also given that m<ABC = 35 degrees.
c)Using substitution property of equality, you have (35°+ m<CBD = 90 degrees.)
If we subtract 35 from both sides, we get 35-35+ m<CBD = 90-35.
Therefore, m<CBD = 55 degrees.
PLEASE CHECK THE ATTACHED FIGURE TO SEE HOW THE QUESTION LOOKS LIKE
First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of
possible passwords.
Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of
possible passwords.
So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is
Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have
As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:
Where x can be any of the 36 characters.
So, we have 25 cases with 4 vacant slots, leading to a probability of
Finally, you can compute the probability of the union using the formula
Since we already computed all these quantities.