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olga nikolaevna [1]
3 years ago
6

What is the probability of rolling a 4 on a single sided die and then rolling a 6?

Mathematics
1 answer:
Rzqust [24]3 years ago
4 0

Answer:

If it is a six-sided die then it has the probability of 1/6 or .16

Step-by-step explanation:

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What is the solution set for
Mumz [18]

Answer: C. (-3, 9)

Step-by-step explanation:

Given

2|x-3|=12

Divide both sides by 2

2|x-3|/2=12/2

|x-3|=6

Take of absolute value sign (REMEMBER need to take into consideration of two conditions)

x-3=±6

1)x-3=6

     x=9

2)x-3=-6

      x=-3

**NOTE**: need to check each solution whether or not they are suitable

Hope this helps!! :)

Please let me know if you have any questions

5 0
3 years ago
Marcos delivers newspapers during his summer break. He delivers a different number of papers each day. What percent of the total
soldier1979 [14.2K]
Is there some sort of graph or second part to this?
7 0
3 years ago
Read 2 more answers
Plz help this is from Khan academy ty!!
Trava [24]

Answer:

The answer is A

7 0
2 years ago
The library in Richmond collected $4,170 in overdue fees last year. This year, the library collected $6,672. What is the percent
melamori03 [73]

Answer:

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Step-by-step explanation:

4 0
2 years ago
Find the minimum sample size needed to estimate the percentage of Democrats who have a sibling. Use a 0.1 margin of error, use a
Brums [2.3K]

Answer:

The minimum sample size is  n  =135

Step-by-step explanation:

From the question we are told that

   The confidence interval is ( lower \ limit  = \  0.44,\ \ \   upper \ limit  = \ 0.51)

    The margin of error is  E =  0.1

   

Generally the sample  proportion can be mathematically evaluated as

     \r p = \frac{ upper \ limit  + lower \ limit }{2}

    \r p = \frac{ 0.51 + 0.44}{2}

    \r p = 0.475

Given that the confidence level is  98% then the level of significance can be mathematically evaluated as

         \alpha = 100 -  98

        \alpha = 2\%

        \alpha =0.02

Next we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table  

   The value is

        Z_{\frac{\alpha }{2} } =  2.33

Generally the minimum sample size is evaluated as

      n  =[ \frac { Z_{\frac{\alpha }{2} }}{E} ]^2 *  \r p (1- \r p )

     n  =[ \frac { 2.33}{0.1} ]^2 *  0.475(1- 0.475 )

     n  =135

6 0
3 years ago
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