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Dennis_Churaev [7]
3 years ago
9

Help me ASAP plz!!!!

Mathematics
2 answers:
elena55 [62]3 years ago
6 0
(2, -1), (6,- 4), and (6, -1). Hope this helps man!
puteri [66]3 years ago
5 0
(2, -1), (6,- 1), and (6, -4). 
You might be interested in
Find the probability of rolling an odd sum or a sum less than 7 when a pair of dice is rolled
Nutka1998 [239]

Answer:

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

Step-by-step explanation:

For this case when a pair of dice is rolled we have the following sample pace for the outcomes:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

We see 36 possible outcomes and we want to find how many of these pairs we got rolling an odd sum or a sum less than 7

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

7 0
3 years ago
SOLVE THIS PROBLEM ASAP PLS
deff fn [24]

Answer:

See attached

Step-by-step explanation:

-> Also see attached

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

7 0
2 years ago
Jack’s oak tree is 10 feet tall and grows
dalvyx [7]

Answer: 0 months its already shorter than hanks tree

8 0
3 years ago
Read 2 more answers
Plz help me I am trying hard i have been in here for an hour plz help me
amm1812

Answer:

1: 6 ^ 3

2: Neither

3: 6 ^ 3

4: 6 ^ -3

Step-by-step explanation:

First off, GOOD LUCK IXL IS AHDIOFJDSFS

(Second, I hope I am correct and gotchu, I think I am though, have a nice day!)

<u>Third:</u>

1: 6 ^ 3 If you mutiple it out then the negatives will cancel each other out

2: Right off the bat we can tell it is neither because a 3 times a 3 won't equal a 3.

3: 6 ^ 3 (flip fraction and change sign)

4: 6 ^ -3 (flip fraction and change sign)

4 0
2 years ago
Read 2 more answers
Nadine and Calvin are simplifying the expression (StartFraction r Superscript negative 5 Baseline s Superscript negative 3 Basel
Nat2105 [25]

Answer:

Calvin's first step is to simplify the expression is to apply the quotient of powers to get (r Superscript negative 13 Baseline s Superscript negative 1 Baseline) Superscript negative 4 is the correct step

That is (\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}=(r^{-13}s^{-1})^{-4} Calvin's step is the correct step.Because this is the correct way to do simplify the rational expression. And also because Nadine made a blender mistake in her operations in step

Step-by-step explanation:

Given that Nadine and Calvin are simplifying the expression (StartFraction r Superscript negative 5 Baseline s Superscript negative 3 Baseline Over r Superscript 8 Baseline s Superscript negative 2 Baseline EndFraction) Superscript negative 4

Their expression can be written as below

(\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}

Nadine's first step is to simplify the expression is to raise the numerator and denominator to the power of 4 to get StartFraction r Superscript negative 20 Baseline s Superscript negative 12 Baseline Over r Superscript 32 Baseline s Superscript 8 Baseline EndFraction

That is \frac{r^{20}s^{12}}{r^{-32}s^8}

Calvin's first step is to simplify the expression is to apply the quotient of powers to get (r Superscript negative 13 Baseline s Superscript negative 1 Baseline) Superscript negative 4

That is r^{-13}s^{-1}

Now simplify the given expression to check whose step is correct:

(\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}

=(r^{-5}s^{-3}r^{-8}s^{2})^{-4} ( using the property \frac{1}{a^m}=a^{-m} )

=(r^{-5-8}s^{-3+2})^{-4}

=(r^{-13}s^{-1})^{-4}

Therefore (\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}=(r^{-13}s^{-1})^{-4}

Therefore Calvin's first step is to simplify the expression is to apply the quotient of powers to get (r Superscript negative 13 Baseline s Superscript negative 1 Baseline) Superscript negative 4 is the correct step.

That is (\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}=(r^{-13}s^{-1})^{-4} Calvin's step is the correct step .Because this is the correct way to do simplify the rational expression.And also because Nadine made a blender mistake in her operations in step

6 0
2 years ago
Read 2 more answers
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