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antiseptic1488 [7]
3 years ago
13

.What is the probability of two 6 sided diced rolled, adding up to lucky 13

Mathematics
1 answer:
BARSIC [14]3 years ago
5 0

Answer:

0

Step-by-step explanation:

Assuming the six sided dice are numbered 1 through six

The max each could roll is 6

6+6 = 12

They cannot reach a sum of 13

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What is log base 5(4*7 )+log base 5 of 2 written as a single log?
nataly862011 [7]
The answer for the exercise shown above is the last option (Option D), which is:

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The explanation is show below:

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 <span>log5(4*7 )+log5(2)
</span>
 2. By the logarithms properties, you can rewrite the logarithm expression as following:

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 3. Therefore, as you can see, the answer is the option mention before.
 
7 0
3 years ago
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Half the sum of x and you decreased by one-third of y
geniusboy [140]

Answer:

½(x+y) - y/3. Hope this helps! If not, sorry!

Step-by-step explanation:


4 0
3 years ago
In which of the following situations would the use of sampling be most appropriate?Multiple Choice
pishuonlain [190]

Answer:

The use of sampling would be best in the following situation:

a. The need for precise information is less important.

Step-by-step explanation:

Sampling:

It is such a process of analysis in which we divide a large proportion of data into smaller proportions called samples to determine the characteristics of that data.

  • The option a is correct as in sampling, we take a smaller proportion from a large pool of data so when the precise information is less important, it is a good way to use sampling.
  • The option b is not correct as the number of items comprising the population is always large.
  • The option c is not correct as the likelihood of selecting a representative is relatively not small rather it is large.
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5 0
3 years ago
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Feliz [49]

Answer:

The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).

Step-by-step explanation:

We have a small sample size n = 25, \bar{x} = 93.36 and s = 19.95. The confidence interval is given by  \bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}}) where t_{\alpha/2} is the \alpha/2th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that \alpha = 0.02 and the confidence interval is 93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}}) where t_{0.01} is the 1st quantile of the t distribution with 24 df, i.e., t_{0.01} = -2.4922. Then, we have 93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}}) and the 98% confidence interval is given by (83.4161, 103.3039).

3 0
3 years ago
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