Leanne is picking out some movies to rent, and she is primarily interested in children's movies and comedies. She has narrowed down her selections to 18 children's movies and 7 comedies. Step 2 of 2 : How many different combinations of 3 movies can she rent if she wants at least one comedy

If she wants at least one comedy, there are 1484 different combinations.

Step-by-step explanation:

The order in which she wants to pick the movies is not important. So, we use the combinations formula to solve this question.

Combinations formula:

is the number of different combinations of x objects from a set of n elements, given by the following formula.

In this question:

She wants combinations of 3 movies, with at least one comedy. The easiest way to find this is finding the total number of combinations of 3 movies, from the set of 25(18 children's and 7 comedies), and subtract by the total number without comedies(which is 3 from a set of 25). So

Total:

3 from a set of 25.

Without comedies:

3 from a set of 18.

At least one comedy:

If she wants at least one comedy, there are 1484 different combinations.

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.