The number of matinee movies attended is 4.
The number of a evening show movies attended is 2.
<u>Step-by-step explanation:</u>
- Let x represent the number of matinee movies attended.
- Let y represent the number of evening show movies attended.
- Alejandro went to see a total of 6 movies.
Therefore, from the given data the equation can be framed as :
⇒ x + y = 6 ----------(1)
- The cost of a matinee is $7.
- The cost of an evening show is $12.
- Alejandro spent a total of $52.
Therefore, from the given data the equation can be framed as :
⇒ 7x + `12y = 52 ---------(2)
<u>To solve the equations for x and y values :</u>
Mulitply eq (1) and by 7 and subtract eq (2) from eq (1),
7x + 7y = 42
- <u>(7x + 12y = 52)</u>
<u> - 5y = - 10 </u>
⇒ y = 10/5
⇒ y = 2
The number of a evening show movies attended is 2.
Substitute y=2 in eq (1),
⇒ x+2 = 6
⇒ x = 6-2
⇒ x = 4
The number of matinee movies attended is 4.
Answer:
D. H = 2.1t + 12
Step-by-step explanation:
Let us assume the t be the time in minutes
And h be the water level
Now before reaching to the equation that need to be used
before this, determine the inches per minute i.e.
= 54 inches - 12 inches
= 42 inches in 20 minutes.
Now for per minute, it would be
= 42 inches ÷ 20 minutes
= 2.1 inches per minute
After this, multiply the inches per minute with the minutes t and add the sale to the beginning value i.e. 12
So the equation is
H = 2.1t + 12
hence, the last option is correct
Answer:15/17
Step-by-step explanation:
Ask your teacher that’s all you have to do.
Answer: P(odd) = 0.499
Step-by-step explanation:
Given:
Total number of people = 20
Number of men = 12
Number of women = 8
Number of jury to be selected = 6
For the jury to have an odd number of women. it must have either of the three.
1. 1 woman , 5 men
2. 3 women, 3 men
3. 5 women, 1 man
The total possible ways of selecting the 6 people jury is;
N = 20C6 = 20!/6!(20-6)!
N = 38760
The possible ways of selecting;
Case 1 : 1 woman, 5 men
N1 = 8C1 × 12C5
N1 = 8 × 792 = 6336
Case 2 : 3 women , 3 men
N2 = 8C3 × 12C3
N2 = 12320
Case 3 : 5 women, 1 man
N3 = 8C5 × 12C1
N3 = 672
P(Odd) = (N1+N2+N3)/N
P(odd) = (6336+12320+672)/38760
P(odd) = 19328/38760
P(odd) = 0.499