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AleksAgata [21]
2 years ago
11

John wants to predict the frequency of rolling a 4 or 5 using a 6-sided dice with numbers one through six. Which of the followin

g is the closest to the prediction if the number cube was rolled 600 times?
A: about 200 times
B: about 120 times
C: about 100 times
D: about 16 times
Mathematics
1 answer:
weeeeeb [17]2 years ago
4 0

Answer: the answer is 120 or b

Step-by-step explanation: the answer is b

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If you know the answer your a genius at math^
nirvana33 [79]

Answer: I think its B

Step-by-step explanation: hoped this helped :)

4 0
2 years ago
Arthur bought a life insurance policy at $10.98 per month for a 20 year term. What will he pay over 20 years for the premium?
andre [41]
12(months) x 20 (years) = 240

10.98 x 240 = 2635.20

So Arthur will pay $2635.20 over 20 years for premium.

I hope this helps :)
5 0
2 years ago
Read 2 more answers
Let A, B, C and D be sets. Prove that A \ B and C \ D are disjoint if and only if A ∩ C ⊆ B ∪ D
ANEK [815]

Step-by-step explanation:

We have to prove both implications of the affirmation.

1) Let's assume that A \ B and C \ D are disjoint, we have to prove that A ∩ C ⊆ B ∪ D.

We'll prove it by reducing to absurd.

Let's suppose that A ∩ C ⊄ B ∪ D. That means that there is an element x that belongs to A ∩ C but not to B ∪ D.

As x belongs to A ∩ C, x ∈ A and x ∈ C.

As x doesn't belong to B ∪ D, x ∉ B and x ∉ D.

With this, we can say that x ∈ A \ B and x ∈ C \ D.

Therefore, x ∈ (A \ B) ∩ (C \ D), absurd!

It's absurd because we were assuming that A \ B and C \ D were disjoint, therefore their intersection must be empty.

The absurd came from assuming that A ∩ C ⊄ B ∪ D.

That proves that A ∩ C ⊆ B ∪ D.

2) Let's assume that A ∩ C ⊆ B ∪ D, we have to prove that A \ B and C \ D are disjoint (i.e.  A \ B ∩ C \ D is empty)

We'll prove it again by reducing to absurd.

Let's suppose that  A \ B ∩ C \ D is not empty. That means there is an element x that belongs to  A \ B ∩ C \ D. Therefore, x ∈ A \ B and x ∈ C \ D.

As x ∈ A \ B, x belongs to A but x doesn't belong to B.  

As x ∈ C \ D, x belongs to C but x doesn't belong to D.

With this, we can say that x ∈ A ∩ C and x ∉ B ∪ D.

So, there is an element that belongs to A ∩ C but not to B∪D, absurd!

It's absurd because we were assuming that A ∩ C ⊆ B ∪ D, therefore every element of A ∩ C must belong to B ∪ D.

The absurd came from assuming that A \ B ∩ C \ D is not empty.

That proves that A \ B ∩ C \ D is empty, i.e. A \ B and C \ D are disjoint.

7 0
3 years ago
1.29 divided by 1.25 =
Ronch [10]
Answer: 1.032 

1.29/1.25 = 1.032
5 0
3 years ago
Read 2 more answers
I will attach a screenshot of the math problem.
Alekssandra [29.7K]

Complete the table for the function y = 0.1^x

The first step: plug values from the left column into the ‘x’ spot in the formula <u>y=0.1^x</u>.

* 0.1^-2 : We can eliminate the negative exponent value by using the rule a^-1 = 1/a. Keep this rule in mind for future problems. (0.1^-2 = 1/0.1 * 0.1 = 100).

* 0.1^-1 = 1/0.1 = 10

* 0.1^0 = 1 : (Remember this rule: a^0 = 1)

* 0.1^1 = 0.1

Our list of values: 100, 10, 1, 0.1

Now, we can plug these values into your table:

\left[\begin{array}{ccc}x&y\\2&10\\1&10\\0&1\\1&0.1\end{array}\right]

The points can now be graphed. I will paste a Desmos screenshot; try to see if you can find some of the indicated (x,y) values: [screenshot is attached]

I hope this helped!

7 0
2 years ago
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