PL JUST HELPPP MY IM CRYINGGG HELP
At a fair, each person can spin two wheels of chance. The first wheel has the numbers 1 and 4. The second
wheel has the letters M, A, T, H.
List all the possible outcomes of the
compound event.
What is the probability of spinning a 1 on
the first wheel?
What is the probability of spinning a T on
the second wheel?
If you spin both wheels, what is the
probability that you get either a 1 or a T?
If you spin both wheels, what is the
probability of getting a 1 and a T?

Please don't cry.  There's no crying in math.

Wheel 1:    1 2 3 4     Wheel 2:  M A T H

Compound Outcomes: 1M 1A 1T 1H 2M 2A 2T 2H 3M 3A 3T 3H 4M 4A 4T 4H

A 1 on wheel 1 is one of four equally likely possibilities.

P(Wheel1=1) = 1/4

There's one T out of 4 letters on wheel 2.

P(Wheel2=T) = 1/4

Just going through our list of 16, I count 7 with a 1 or T, namely 1M 1A 1T 1H 2T 3T 4T.  So

P(1 or T) = 7/16

There's only one way to get a 1 and a T out of the sixteen, namely 1T.

P(1 and T) = 1/16

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844

∠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.