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Margarita [4]
3 years ago
7

Emmie rolls a dice three times. What is the probability that she will only roll numbers greater than two on all of her rolls?

Mathematics
1 answer:
Likurg_2 [28]3 years ago
6 0
12/18 or 2/3 chance simplified
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A father is twice as old as his son and the sum of their ages is 75. How old is the<br> father?
castortr0y [4]

Answer: 50

Let's first of all assume the son's age as x.

So,

Father's age will be 2x i.e. twice as old as the son

Given that,

Sum of their ages = 75 years

So,

x + 2x = 75

3x = 75

x = 75/3

x = 25

Hence,

Son's age = x = 25 years

Father's age = 2x = 2 * 25 = 50 years

8 0
2 years ago
A 28<br> B42<br> C47<br> D 33<br><br> Please help
netineya [11]
D.) There are always 180 degrees in triangles. 147 is already shown, so subtract that from 180. That gives you 33.
5 0
3 years ago
Add the two expressions 2x + 6 + 6x - 1​
castortr0y [4]

Answer: 2x+6+6x-1  =  8x+5

5 0
3 years ago
Read 2 more answers
Mislabeled seafood In 2013 the environmental group Oceana (usa.oceana.org) analyzed 1215 samples of seafood purchased across the
Step2247 [10]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

<h3>z-distribution interval:</h3>

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

  • In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

For this problem:

  • 1215 samples, hence n = 1215.
  • 33% was mislabeled or misidentified, hence p = 0.33.
  • 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

<h3>The lower limit of this interval is:</h3>

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 - 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3036

<h3>The upper limit of this interval is:</h3>

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 + 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3564

The 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

You can learn more about the use of the z-distribution to build a confidence interval at brainly.com/question/25730047

4 0
2 years ago
2. Write 1.68 as a simplified fraction
Alex787 [66]
  • <em><u>1  \frac{68}{100 }</u></em>
5 0
3 years ago
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